MCQ 11 Mark
If $n$ is a positive integer, then ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^{4n + 1}}$=
Answerc
(c)Since $\frac{{1 + i}}{{1 - i}} = \frac{{(1 + i)(1 + i)}}{{(1 - i)(1 + i)}} = i$
Therefore ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^{4n + 1}} = {i^{4n + 1}} = i{i^{4n}} = i\,\,\,\,\,({i^{4n}} = 1)$.
View full question & answer→MCQ 21 Mark
If $(1 - i)^n = 2^n,$ then $n = $
Answerb
(b) If $(1 - i)^n = 2^n$ ......$(i)$
We know that if two complex numbers are equal, their moduli must also be equal, therefore from $(i)$, we have
$|(1 - i)^n|\, = \,|2^n|$
$ \Rightarrow $ $|1 - i|^n = \,|2|^n$,$(\because \,\,2^n > 0)$
==> $\left[ \sqrt {{1^2} + {{( - 1)}^2}} \right]^n = 2^n$
==> $(\sqrt 2 )^n = 2^n$
==> $2^{n/2} = 2^n$
==> $\frac{n}{2} = n$
==>$n = 0$
Trick : By inspection, ${(1 - i)^0} = {2^0}\,\,\,\, \Rightarrow 1 = 1$.
View full question & answer→MCQ 31 Mark
The value of $\frac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = $
Answerb
(b) $\frac{{{i^{584}}({i^8} + {i^6} + {i^4} + {i^2} + 1)}}{{{i^{574}}({i^8} + {i^6} + {i^4} + {i^2} + 1)}} - 1 = \frac{{{i^{584}}}}{{{i^{574}}}} - 1$
$ = {i^{10}} - 1 = - 1 - 1 = - 2$.
View full question & answer→MCQ 41 Mark
$1 + {i^2} + {i^4} + {i^6} + ..... + {i^{2n}}$is
Answerd
(d) $S = 1 + {i^2} + {i^4} + ..... + {i^{2n}} = 1 - 1 + 1 - 1 + ...... + {( - 1)^n}$
Obviously it depends on $n$.
Hence cannot be determined unless $n$ is known.
View full question & answer→MCQ 51 Mark
The value of ${i^{1 + 3 + 5 + ... + (2n + 1)}}$ is
- A
$i$ if $n$ is even, $-i$ if $n$ is odd
- B
$1$ if $n$ is even, $-1$ if $n$ is odd
- ✓
$1$ if $n$ is odd, $-1$ if $n$ is even
- D
$i$ if $n$ is even,$ -1$ if $n$ is odd
AnswerCorrect option: C. $1$ if $n$ is odd, $-1$ if $n$ is even
c
(c) Let $z = i^{[1 + 3 + 5 + .... + (2n + 1)]}$
Clearly series is A.P. with common difference $= 2$
$\because \,T_n = 2n - 1$and ${T_{n + 1}} = 2n + 1$
So, number of terms in A. P. $ = n + 1$
Now, ${S_{n + 1}} = \frac{{n + 1}}{2}[2.1 + (n + 1 - 1)2]$
$ \Rightarrow {S_{n + 1}} = \frac{{n + 1}}{2}[2 + 2n] = (n + 1)^2$ i.e. $i^{(n + 1)^2}$
Now put $n = 1,\,2,\,3,\,4,\,5,\,.....$
$n = 1,z = {i^4} = 1$, $n = 2,\,z = {i^6} = - 1$,
$n = 3,\,z = {i^8} = 1$, $n = 4,\,z = {i^{10}} = - 1$,
$n = 5,\,\,z = {i^{12}} = 1\,,........$
View full question & answer→MCQ 61 Mark
$\left( {\frac{1}{{1 - 2i}} + \frac{3}{{1 + i}}} \right)\,\,\left( {\frac{{3 + 4i}}{{2 - 4i}}} \right) = $
- A
$\frac{1}{2} + \frac{9}{2}i$
- B
$\frac{1}{2} - \frac{9}{2}i$
- C
$\frac{1}{4} - \frac{9}{4}i$
- ✓
$\frac{1}{4} + \frac{9}{4}i$
AnswerCorrect option: D. $\frac{1}{4} + \frac{9}{4}i$
d
(d)$\left( {\frac{1}{{1 - 2i}} + \frac{3}{{1 + i}}} \right)\,\,\left( {\frac{{3 + 4i}}{{2 - 4i}}} \right)$
$ = \left[ {\frac{{1 + 2i}}{{{1^2} + {2^2}}} + \frac{{3 - 3i}}{{{1^2} + {1^2}}}} \right]\,\left[ {\frac{{6 - 16 + 12i + 8i}}{{{2^2} + {4^2}}}} \right]$
$ = \left( {\frac{{2 + 4i + 15 - 15i}}{{10}}} \right)\,\,\left( {\frac{{ - 1 + 2i}}{2}} \right)$
$ = \frac{{(17 - 11i)( - 1 + 2i)}}{{20}} = \frac{{5 + 45i}}{{20}} = \frac{1}{4} + \frac{9}{4}i$.
View full question & answer→MCQ 71 Mark
$\frac{{\sqrt {5 + 12i} + \sqrt {5 - 12i} }}{{\sqrt {5 + 12i} - \sqrt {5 - 12i} }} = $
- ✓
$ - \frac{3}{2}i$
- B
$\frac{3}{2}i$
- C
$ - \frac{3}{2}$
- D
$\frac{3}{2}$
AnswerCorrect option: A. $ - \frac{3}{2}i$
a
(a) $\frac{{(\sqrt {5 + 12i} + \sqrt {5 - 12i} )(\sqrt {5 + 12i} + \sqrt {5 - 12i} )}}{{(\sqrt {5 + 12i} - \sqrt {5 - 12i} )(\sqrt {5 + 12i} + \sqrt {5 - 12i} )}}$
$ = \frac{{5 + 12i + 5 - 12i + 2\sqrt {5 + 12i} \sqrt {5 - 12i} }}{{5 + 12i - 5 + 12i}}$
$ = \frac{{10 + 2 \times ( \pm 13)}}{{24i}} = - \frac{3}{2}i$ or $\frac{{2i}}{3}$.
View full question & answer→MCQ 81 Mark
If ${a^2} + {b^2} = 1,$ then $\frac{{1 + b + ia}}{{1 + b - ia}} = $
- A
$1$
- B
$2$
- ✓
$b + ia$
- D
$a + ib$
AnswerCorrect option: C. $b + ia$
c
(c) Given that ${a^2} + {b^2} = 1$, therefore
$\frac{{1 + b + ia}}{{1 + b - ia}} = \frac{{(1 + b + ia)(1 + b + ia)}}{{(1 + b - ia)(1 + b + ia)}}$
$ = \frac{{{{(1 + b)}^2} - {a^2} + 2ia(1 + b)}}{{1 + {b^2} + 2b + {a^2}}}$$ = \frac{{(1 - {a^2}) + 2b + {b^2} + 2ia(1 + b)}}{{2(1 + b)}}$
$ = \frac{{2{b^2} + 2b + 2ia(1 + b)}}{{2\,(1 + b)}} = b + ia$
Trick : Put $a = 0,b = 1$, $\frac{{1 + b + ia}}{{1 + b - ia}} = \frac{{1 + 1 + 0}}{{1 + 1 - 0}} = 1$
But options $(a)$ and $ (c)$ give $1.$
So again put $a = 1,b = 0,\frac{{1 + b + ia}}{{1 + b - ia}} = \frac{{1 + i}}{{1 - i}} = i$.
Which gives $ (c)$ only.
View full question & answer→MCQ 91 Mark
If ${(x + iy)^{1/3}} = a + ib,$then $\frac{x}{a} + \frac{y}{b}$is equal to
- A
$4({a^2} + {b^2})$
- ✓
$4({a^2} - {b^2})$
- C
$4({b^2} - {a^2})$
- D
AnswerCorrect option: B. $4({a^2} - {b^2})$
b
(b) ${(x + iy)^{1/3}} = a + ib$==>$(x + iy) = {(a + ib)^3}$
$ = {a^3} + 3{a^2}.ib + 3a.{(ib)^2} + {(ib)^3}$
$ = {a^3} - 3a{b^2} + i(3{a^2}b - {b^3})$
Equating real and imaginary parts, we get
$\frac{x}{a} = {a^2} - 3{b^2}$and $\frac{y}{b} = 3{a^2} - {b^2}$
$\therefore $ $\frac{x}{a} + \frac{y}{b} = 4({a^2} - {b^2})$
View full question & answer→MCQ 101 Mark
The real values of $x$ and $y$ for which the equation $({x^4} + 2xi) - (3{x^2} + yi) = $$(3 - 5i) + (1 + 2yi)$ is satisfied, are
AnswerCorrect option: C. Both $(a)$ and $(b)$
c
(c) Given equation
$({x^4} + 2xi) - (3{x^2} + yi) = (3 - 5i) + (1 + 2yi)$
$ \Rightarrow \,\,\,({x^4} - 3{x^2}) + i(2x - 3y) = 4 - 5i$
Equating real and imaginary parts, we get
${x^4} - 3{x^2} = 4$ ......$(i)$
and $2x - 3y = - 5$ .....$(ii)$
From $(i)$ and $(ii),$ we get $x = \pm 2$and $y = 3,\frac{1}{3}$
Trick : Put $x = 2,y = 3$and then $x = - 2,$$y = \frac{1}{3},$ we see that they both satisfy the given equation.
View full question & answer→MCQ 111 Mark
If $x + iy = \frac{3}{{2 + \cos \theta + i\sin \theta }},$then ${x^2} + {y^2}$ is equal to
- A
$3x - 4$
- ✓
$4x - 3$
- C
$4x + 3$
- D
AnswerCorrect option: B. $4x - 3$
b
(b) If $x + iy = \frac{3}{{2 + \cos \theta + i\sin \theta }}$
$ = \frac{{3(2 + \cos \theta - i\sin \theta )}}{{{{(2 + \cos \theta )}^2} + {{\sin }^2}\theta }} = \frac{{6 + 3\cos \theta - 3i\sin \theta }}{{4 + {{\cos }^2}\theta + 4\cos \theta + {{\sin }^2}\theta }}$
$ = \left[ {\frac{{6 + 3\cos \theta }}{{5 + 4\cos \theta }}} \right] + i\,\left[ {\frac{{ - 3\sin \theta }}{{5 + 4\cos \theta }}} \right]$
==> $x = \frac{{3(2 + \cos \theta )}}{{5 + 4\cos \theta }},y = \frac{{ - 3\sin \theta }}{{5 + 4\cos \theta }}$
${x^2} + {y^2} = \frac{9}{{{{(5 + 4\cos \theta )}^2}}}$ $[4 + {\cos ^2}\theta + 4\cos \theta + {\sin ^2}\theta ]$
$ = \frac{9}{{5 + 4\cos \theta }} = 4\left[ {\frac{{6 + 3\cos \theta }}{{5 + 4\cos \theta }}} \right] - 3 = 4x - 3$
Trick : $x + iy = \frac{{3(2 + \cos \theta - i\sin \theta )}}{{(2 + \cos \theta + i\sin \theta )(2 + \cos \theta - i\sin \theta )}}$
Let $y = 0$, then $\sin \theta = 0$i.e., $\theta = 0$.
Now put $x = 1$ then ${x^2} + {y^2} = {1^2} + 0 = 1$.
Also option (b) gives $4(1) -3=1.$
View full question & answer→MCQ 121 Mark
If $\frac{{{{(p + i)}^2}}}{{2p - i}} = \mu + i\lambda ,$then ${\mu ^2} + {\lambda ^2}$ is equal to
- A
$\frac{{{{({p^2} + 1)}^2}}}{{4{p^2} - 1}}$
- B
$\frac{{{{({p^2} - 1)}^2}}}{{4{p^2} - 1}}$
- C
$\frac{{{{({p^2} - 1)}^2}}}{{4{p^2} + 1}}$
- ✓
$\frac{{{{({p^2} + 1)}^2}}}{{4{p^2} + 1}}$
AnswerCorrect option: D. $\frac{{{{({p^2} + 1)}^2}}}{{4{p^2} + 1}}$
d
(d) $\mu + i\lambda = \frac{{{{(p + i)}^2}}}{{2p - i}} = \frac{{({p^2} - 1 + 2pi)(2p + i)}}{{(2p - i)(2p + i)}}$
$ = \frac{{2p({p^2} - 2) + i(5{p^2} - 1)}}{{4{p^2} + 1}}$
${\mu ^2} + {\lambda ^2} = \frac{{4{p^2}{{({p^2} - 2)}^2} + {{(5{p^2} - 1)}^2}}}{{{{(4{p^2} + 1)}^2}}}$
$ = \frac{{4{p^6} + 6{p^2} + 9{p^4} + 1}}{{{{(4{p^2} + 1)}^2}}}$
$ = \,\,\frac{{{p^4}(4{p^2} + 1) + 2{p^2}(4{p^2} + 1) + (4{p^2} + 1)}}{{{{(4{p^2} + 1)}^2}}}$
$ = \frac{{{p^4} + 2{p^2} + 1}}{{4{p^2} + 1}} = \frac{{{{({p^2} + 1)}^2}}}{{4{p^2} + 1}}$.
View full question & answer→MCQ 131 Mark
If $\frac{{3x + 2iy}}{{5i - 2}} = \frac{{15}}{{8x + 3iy}}$, then
AnswerCorrect option: D. $x = - 1,y = - 3$or $x = 1,$$y = 3$
d
(d) Given that $\frac{{3x + 2iy}}{{5i - 2}} = \frac{{15}}{{8x + 3iy}}$
==> $24{x^2} + 9ixy - 6{y^2} + 16ixy = 75i - 30$
==> $24{x^2} - 6{y^2} + 25ixy = 75i - 30$
Equating real and imaginary parts, we get
$24{x^2} - 6{y^2} = - 30$or $4{x^2} - {y^2} = - 5$and $xy = 3$
On solving we get $x = \pm 1,y = \pm 3$
View full question & answer→MCQ 141 Mark
If $z(1 + a) = b + ic$ and ${a^2} + {b^2} + {c^2} = 1$, then $\frac{{1 + iz}}{{1 - iz}} = $
- ✓
$\frac{{a + ib}}{{1 + c}}$
- B
$\frac{{b - ic}}{{1 + a}}$
- C
$\frac{{a + ic}}{{1 + b}}$
- D
AnswerCorrect option: A. $\frac{{a + ib}}{{1 + c}}$
a
(a) $\frac{{1 + iz}}{{1 - iz}} = \frac{{1 + i(b + ic)/(1 + a)}}{{1 - i(b + ic)/(1 + a)}} = \frac{{1 + a - c + ib}}{{1 + a + c - ib}}$
$ = \frac{{(1 + a - c + ib)(1 + a + c + ib)}}{{{{(1 + a + c)}^2} + {b^2}}}$
$ = \frac{{1 + 2a + {a^2} - {b^2} - {c^2} + 2ib + 2iab)}}{{1 + {a^2} + {c^2} + {b^2} + 2ac + 2(a + c)}}$
= $\frac{{{a^2} + {b^2} + {c^2} + 2a + {a^2} - {b^2} - {c^2} + 2ib(1 + a)}}{{1 + 1 + 2ac + 2(a + c)}}$
$ = \frac{{2a(a + 1) + 2ib(1 + a)}}{{2(1 + a)(1 + c)}} = \frac{{a + ib}}{{1 + c}}$.
View full question & answer→MCQ 151 Mark
Let ${z_1},{z_2}$ be two complex numbers such that ${z_1} + {z_2}$ and ${z_1}{z_2}$ both are real, then
- A
${z_1} = - {z_2}$
- ✓
${z_1} = {\bar z_2}$
- C
${z_1} = - {\bar z_2}$
- D
${z_1} = {z_2}$
AnswerCorrect option: B. ${z_1} = {\bar z_2}$
b
(b) Let ${z_1} = a + ib,{z_2} = c + id$, then
${z_1} + {z_2}$ is real ==> $(a + c) + i(b + d)$is real
==> $b + d = 0$ ==> $d = - b$ .....(i)
${z_1}{z_2}$ is real ==> $(ad - bd) + i(ac + bc)$is real
==> $ad + bc = 0$ ==> $a( - b) + bc = 0$==> $a = c$
${z_1} = a + ib = c - id = {\bar z_2}$ $(\because a = c$ and $b = - d)$
View full question & answer→MCQ 161 Mark
$A + iB$ form of $\frac{{(\cos x + i\sin x)(\cos y + i\sin y)}}{{(\cot u + i)(1 + i\tan v)}}$ is
- ✓
$\sin u\cos v\,[\cos (x + y - u - v) + i\sin (x + y - u - v)]$
- B
$\sin u\cos v\,[\cos (x + y + u + v) + i\sin (x + y + u + v)]$
- C
$\sin u\cos v\,[\cos (x + y + u + v) - i\sin (x + y + u + v)]$
- D
AnswerCorrect option: A. $\sin u\cos v\,[\cos (x + y - u - v) + i\sin (x + y - u - v)]$
a
(a)$L.H.S.$ $ = \frac{{(\cos x + i\sin x)(\cos y + i\sin y)}}{{(\cos u + i\sin u)(\cos v + i\sin v)}}$$\sin u\cos v$
$ = \sin u\cos v[\cos (x + y - u - v) + i\sin (x + y - u - v)]$
View full question & answer→MCQ 171 Mark
If $x,y \in R$and $(x + iy)(3 + 2i) = 1 + i$, then $(x,\,y)$ is
- A
$\left( {1,\frac{1}{5}} \right)$
- B
$\left( {\frac{1}{{13}},\frac{1}{{13}}} \right)$
- ✓
$\left( {\frac{5}{{13}},\frac{1}{{13}}} \right)$
- D
$\left( {\frac{1}{5},\frac{1}{5}} \right)$
AnswerCorrect option: C. $\left( {\frac{5}{{13}},\frac{1}{{13}}} \right)$
c
(c) $x + iy = \left( {\frac{{1 + i}}{{3 + 2i}}} \right)\frac{{(3 - 2i)}}{{(3 - 2i)}} = \frac{{5 + i}}{{13}}$
Hence $x = \frac{5}{{13}},y = \frac{1}{{13}}$.
View full question & answer→MCQ 181 Mark
If $a = \cos \,\theta + i\,\sin \,\theta ,$ then $\frac{{1 + a}}{{1 - a}} = $
- A
$\cot \theta $
- B
$\cot \frac{\theta }{2}$
- ✓
$i\,\cot \frac{\theta }{2}$
- D
$i\,\tan \frac{\theta }{2}$
AnswerCorrect option: C. $i\,\cot \frac{\theta }{2}$
c
(c) $a = \cos \theta + i\sin \theta .$
$\frac{{1 + a}}{{1 - a}} = \frac{{(1 + \cos \theta ) + i\sin \theta }}{{(1 - \cos \theta ) - i\sin \theta }}.\,$
Rationalization of denominator, we get $\frac{{1 + a}}{{1 - a}} = \frac{{(1 + \cos \theta ) + i\sin \theta }}{{(1 - \cos \theta ) - i\,\sin \theta }} \times \frac{{(1 - \cos \theta ) + i\sin \theta }}{{(1 - \cos \theta ) + i\sin \theta }}$
$ = \frac{{(1 + \cos \theta )\,(1 - \cos \theta ) + (1 + \cos \theta )\,i\sin \theta + (1 - \cos \theta )i\sin \theta + {i^2}{{\sin }^2}\theta }}{{{{(1 - \cos \theta )}^2} - {{(i\sin \theta )}^2}}}$
$ = \frac{{1 - ({{\cos }^2}\theta + {{\sin }^2}\theta ) + 2i\sin \theta }}{{1 + ({{\cos }^2}\theta + {{\sin }^2}\theta ) - 2\,\cos \theta }}$$ = \frac{{2i\sin \theta }}{{2(1 - \cos \theta )}}$
$ = \frac{{i.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\sin }^2}\frac{\theta }{2}}}$$ = i\frac{{\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}} = i\cot \frac{\theta }{2}$.
View full question & answer→MCQ 191 Mark
Solving $3 - 2yi = {9^x} - 7i$, where ${i^2} = - 1,$ for $x$ and $y$ real, we get
- ✓
$x = 0.5\,\,,\,\,y = 3.5$
- B
$x = 5\,\,,\,\,y = 3$
- C
$x = \frac{1}{2}\,\,,\,\,y = 7$
- D
$x = 0,\,y = \frac{{3 + 7i}}{{2i}}$
AnswerCorrect option: A. $x = 0.5\,\,,\,\,y = 3.5$
a
(a) $3 - 2yi = {9^x} - 7i$
Equating real and imaginary parts both sides
${9^x} = 3 \Rightarrow \,{3^{2x}} = {3^1} \Rightarrow 2x = 1 \Rightarrow x = 0.5$
$2y = 7 \Rightarrow \,y = 3.5$.
View full question & answer→MCQ 201 Mark
If $z = x + iy,\,{z^{1/3}} = a - ib$ and $\frac{x}{a} - \frac{y}{b} = k\,({a^2} - {b^2})$ then value of $k$ equals
Answerb
(b)${(x + iy)^{1/3}} = a - ib$
$x + iy = {(a - ib)^3} = ({a^3} - 3a{b^2}) + i({b^3} - 3{a^2}b)$
$⇒$ $x = {a^3} - 3a{b^2},\,y = {b^3} - 3{a^2}b$
$⇒$ $\frac{x}{a} = {a^2} - 3{b^2},\,\frac{y}{b} = {b^2} - 3{a^2}$
$\therefore $ $\frac{x}{a} - \frac{y}{b} = {a^2} - 3{b^2} - {b^2} + 3{a^2}$
$\frac{x}{a} - \frac{y}{b} = 4({a^2} - {b^2}) = k({a^2} - {b^2})$
$\therefore $ $k = 4$.
.
View full question & answer→MCQ 211 Mark
If $(a + ib)(c + id)(e + if)(g + ih)$$ = A + iB,$ then $({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2})$ =
- ✓
${A^2} + {B^2}$
- B
${A^2} - {B^2}$
- C
${A^2}$
- D
${B^2}$
AnswerCorrect option: A. ${A^2} + {B^2}$
a
(a)$(a + ib)(c + id)(e + if)(g + ih) = A + iB$.....$(i)$
==> $(a - ib)(c - id)(e - if)(g - ih) = A - iB$......$(ii)$
Multiplying $(i) $ and $(ii)$, we get
$({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) = {A^2} + {B^2}$
View full question & answer→MCQ 221 Mark
If $z = \frac{{7 - i}}{{3 - 4i}}$ then ${z^{14}} = $
- A
${2^7}$
- B
${2^7}i$
- C
${2^{14}}i$
- ✓
$ - {2^7}i$
AnswerCorrect option: D. $ - {2^7}i$
d
(d) $z = \frac{{7 - i}}{{3 - 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}$=$\frac{{21 + 25i + 4}}{{16 + 9}} = \frac{{25\,(1 + i)}}{{25}}$ = $(1 + i)$
${z^{14}} = {(1 + i)^{14}} = {[{(1 + i)^2}]^7}$= ${(2i)^7} = {2^7}{i^7} = - {2^7}i$.
View full question & answer→MCQ 231 Mark
The number of real values of $a$ satisfying the equation ${a^2} - 2a\sin x + 1 = 0$ is
Answerc
(c)Given equation ${a^2} - 2a\sin x + 1 = 0$
$\therefore $ $a = \frac{{2\sin x \pm \sqrt {4{{\sin }^2}x - 4} }}{2}$ $ = \sin x \pm \sqrt { - (1 - {{\sin }^2}x)} $
$a = \sin x \pm i\cos x$
If $x = \frac{\pi }{2}$==> $a = 1,$$x = {270^o}$==> $a = - 1$.
View full question & answer→MCQ 241 Mark
If $(1 + i)(1 + 2i)(1 + 3i).....(1 + ni) = a + ib$, then $2.5.10....$$(1 + {n^2})$ is equal to
- A
${a^2} - {b^2}$
- ✓
${a^2} + {b^2}$
- C
$\sqrt {{a^2} + {b^2}} $
- D
$\sqrt {{a^2} - {b^2}} $
AnswerCorrect option: B. ${a^2} + {b^2}$
b
(b)We have
$(1 + i)(1 + 2i)(1 + 3i).....(1 + ni) = a + ib$ .....$(i)$
==> $(1 - i)(1 - 2i)(1 - 3i).....(1 - ni) = a - ib$ .....$(ii)$
Multiplying $ (i)$ and $(ii)$, we get $2.5.....(1 + {n^2}) = {a^2} + {b^2}$
View full question & answer→MCQ 251 Mark
The locus of $z$ satisfying the inequality ${\log _{1/3}}|z + 1|\, > $ ${\log _{1/3}}|z - 1|$ is
- ✓
$R\,(z) < 0$
- B
$R\,(z) > 0$
- C
$I\,(z) < 0$
- D
AnswerCorrect option: A. $R\,(z) < 0$
a
(a)We know that ${\log _a}m > {\log _a}n$
==> $m > n$or $m < n$, according as $a > 1$or $0 < a < 1$.
Hence for $z = x + iy$
${\log _{(1/3)}}|z + 1|\, > \,{\log _{(1/3)}}|z - 1| \Rightarrow |z + 1|$$ < \,|z - 1|$
$\left\{ {\because 0 < \frac{1}{3} < 1} \right\}$
==>$|x + iy + 1| < |x + iy - 1|$
==> ${(x + 1)^2} + {y^2} < {(x - 1)^2} + {y^2}$
==> $4x < 0\, \Rightarrow x < 0\,\, \Rightarrow {\mathop{\rm Re}\nolimits} (z) < 0$
View full question & answer→MCQ 261 Mark
Let $\frac{{1 - ix}}{{1 + ix}} = a - ib$ and ${a^2} + {b^2} = 1$, where $a$ and $b$ are real, then $x = $
- A
$\frac{{2a}}{{{{(1 + a)}^2} + {b^2}}}$
- ✓
$\frac{{2b}}{{{{(1 + a)}^2} + {b^2}}}$
- C
$\frac{{2a}}{{{{(1 + b)}^2} + {a^2}}}$
- D
$\frac{{2b}}{{{{(1 + b)}^2} + {a^2}}}$
AnswerCorrect option: B. $\frac{{2b}}{{{{(1 + a)}^2} + {b^2}}}$
b
(b) $\frac{{1 - ix}}{{1 + ix}} = a - ib$
==> $\frac{{(1 - ix)(1 - ix)}}{{(1 + ix)(1 - ix)}} = a - ib$
==> $\frac{{1 - {x^2} - 2ix}}{{1 + {x^2}}} = a - ib$
==> $\frac{{1 - {x^2}}}{{1 + {x^2}}} = a$and $\frac{{2x}}{{1 + {x^2}}} = b$
Now we can write $x$ as $x = \frac{{\frac{{2x}}{{1 + {x^2}}}}}{{\frac{2}{{1 + {x^2}}}}} = \frac{{\frac{{2x}}{{1 + {x^2}}}}}{{\frac{{1 - {x^2}}}{{1 + {x^2}}} + 1}}$
$ = \frac{b}{{1 + a}} = \frac{{2b}}{{1 + 1 + 2a}} = \frac{{2b}}{{1 + ({a^2} + {b^2}) + 2a}} = \frac{{2b}}{{{{(1 + a)}^2} + {b^2}}}$
Trick : $\frac{{1 - ix}}{{1 + ix}} = \frac{{1 - {x^2} - 2ix}}{{1 + {x^2}}} = a - ib$
Let $a = 0$
==> $x = \pm 1$and $b = \pm 1$.
Also option $(b)$ gives $ \pm 1$.
View full question & answer→MCQ 271 Mark
If $|z + 4|\, \le \,3,$ then the greatest and the least value of $|z + 1|$ are
- A
$6, -6$
- ✓
$6, 0$
- C
$7, 2$
- D
$0, -1$
AnswerCorrect option: B. $6, 0$
b
(b) $|z + 4|\, \le 3$==> $ - 3 \le z + 4 \le + 3$
==> $ - 6 \le z + 1 \le 0$==> 0$ \le - (z + 1) \le 6$
==> $0 \le \,|z + 1| \le 6$,
Hence greatest and least values of $|z + 1|$ are $6$ and $0$ respectively.
View full question & answer→MCQ 281 Mark
If $|{a_k}| < 1,{\lambda _k} \ge 0$ for $k = 1,\,2,....n$ and ${\lambda _1} + {\lambda _2} + ... + {\lambda _n} = 1,$ then the value of $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ....{\lambda _n}{a_n}|$ is
Answerd
(d) We have $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ..... + {\lambda _n}{a_n}|$
$ \le |{\lambda _1}{a_1}| + |{\lambda _2}{a_2}| + ..... + |{\lambda _n}{a_n}|$
$ = |{\lambda _1}||{a_1}| + ..... + |{\lambda _n}||{a_n}|$
$ = {\lambda _1}|{a_1}| + ..... + {\lambda _n}|{a_n}|\,$[ each ${\lambda _k}_. \ge 0$]
$ < {\lambda _1} + ..... + {\lambda _n}$
[ $|{a_k}| < $ $1$ and so ${\lambda _k}|{a_k}| < {\lambda _k}$for all $k = 1,2,....n$]
Hence $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ..... + {\lambda _n}{a_n}| < 1$.
Thus $|{\lambda _1}{a_1} + ..... + {\lambda _n}{a_n}| < 1$.
View full question & answer→MCQ 291 Mark
If $z$ is a complex number such that ${z^2} = {(\bar z)^2},$ then
AnswerCorrect option: C. Either $z$ is purely real or purely imaginary
c
(c)Let $z = x + iy$, then its conjugate $\overline z = x - iy$
Given that ${z^2} = {(\overline z )^2}$
==> ${x^2} - {y^2} + 2ixy = {x^2} - {y^2} - 2ixy$==> $4ixy = 0$
If $x \ne 0$ then $y = 0$and if $y \ne 0$then $x = 0$
.
View full question & answer→MCQ 301 Mark
The number of solutions of the equation ${z^2} + \bar z = 0$ is
Answerd
(d) Let $z = x + iy,$ so that $\overline z = x - iy,$ therefore
${z^2} + \overline z = 0\, \Leftrightarrow ({x^2} - {y^2} + x) + i\,(2xy - y) = 0$
Equating real and imaginary parts, we get
${x^2} - {y^2} + x = 0$ .....$(i)$
and $2xy - y = 0$ ==> $y = 0$or $x = \frac{1}{2}$
If $y = 0$, then $(i)$ gives ${x^2} + x = 0\,\, \Rightarrow x = 0$or $x = - 1$
If $x = \frac{1}{2},$ Then ${x^2} - {y^2} + x = 0$
==>${y^2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$==>$y = \pm \frac{{\sqrt 3 }}{2}$
Hence, there are four solutions in all.
View full question & answer→MCQ 311 Mark
If $\frac{{z - i}}{{z + i}}(z \ne - i)$ is a purely imaginary number, then $z.\bar z$ is equal to
Answerb
(b) Here $\frac{{z - i}}{{z + i}} = \frac{{x + i(y - 1)}}{{x + i(y + 1)}}.\frac{{x - i(y + 1)}}{{x - i(y + 1)}}$
$ = \frac{{({x^2} + {y^2} - 1) + i( - 2x)}}{{{x^2} + {{(y + 1)}^2}}}$
As $\frac{{z - i}}{{z + i}}$ is purely imaginary, we get
${x^2} + {y^2} - 1 = 0$==> ${x^2} + {y^2} = 1$==>$z\overline z = 1$.
View full question & answer→MCQ 321 Mark
If $z = 3 + 5i,\,\,{\rm{then }}\,{z^3} + \bar z + 198 = $
- A
$ - 3 - 5i$
- B
$ - 3 + 5i$
- ✓
$3 + 5i$
- D
$3 - 5i$
AnswerCorrect option: C. $3 + 5i$
c
(c) $z = 3 + 5i$, $\bar z = 3 - 5i$
$ \Rightarrow \,$${z^3} = {(3 + 5i)^3} = {3^3} + {(5i)^3} + 3.3.5i\,(3 + 5i)$
$ = - 198 + 10i$
Hence, ${z^3} + \bar z + 198 = 10\,i - 198 + 3 - 5\,i + 198 = 3 + 5\,i$.
View full question & answer→MCQ 331 Mark
If $\frac{{2{z_1}}}{{3{z_2}}}$ is a purely imaginary number, then $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right|$ =
Answerb
(b) As given, let $\frac{{2{z_1}}}{{3{z_2}}} = iy$ or $\frac{{{z_1}}}{{{z_2}}} = \frac{3}{2}iy$, so that
$\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right| = \left| {\frac{{\frac{{{z_1}}}{{{z_2}}} - 1}}{{\frac{{{z_1}}}{{{z_2}}} + 1}}} \right| = \left| {\frac{{\frac{3}{2}iy - 1}}{{\frac{3}{2}iy + 1}}} \right| = \left| {\frac{{1 - \frac{3}{2}iy}}{{1 + \frac{3}{2}iy}}} \right| = 1$
$\left\{ {\because \,\,\,\,|z|\, = \,|\overline z |} \right\}$
View full question & answer→MCQ 341 Mark
If $z$ is a complex number such that $\frac{{z - 1}}{{z + 1}}$ is purely imaginary, then
- A
$|z|\, = 0$
- ✓
$|z|\, = 1$
- C
$|z|\, > 1$
- D
$|z|\, < 1$
AnswerCorrect option: B. $|z|\, = 1$
b
(b) Let $\frac{{z - 1}}{{z + 1}} = iy$ where $y \in R$
This gives $z = \frac{{1 + iy}}{{1 - iy}} = \frac{{1 + iy}}{{1 - iy}} \times \frac{{1 + iy}}{{1 + iy}} = \frac{{(1 - {y^2}) + 2iy}}{{1 + {y^2}}}$
$\therefore $ $|z| = \frac{1}{{1 + {y^2}}}\sqrt {{{(1 - {y^2})}^2} + 4{y^2}} = \frac{{1 + {y^2}}}{{1 + {y^2}}} = 1$.
View full question & answer→MCQ 351 Mark
The solution of the equation $|z| - z = 1 + 2i$ is
- A
$2 - \frac{3}{2}i$
- B
$\frac{3}{2} + 2i$
- ✓
$\frac{3}{2} - 2i$
- D
$ - 2 + \frac{3}{2}i$
AnswerCorrect option: C. $\frac{3}{2} - 2i$
c
(c)$|z| - z = 1 + 2i$
Let $z = x + iy$, therefore $|x + iy| - (x + iy) = 1 + 2i$
Equating real and imaginary parts, we get
$\sqrt {{x^2} + {y^2}} - x = 1$and $y = - 2$==>$x = \frac{3}{2}$
Hence complex number $z = \frac{3}{2} - 2i$.
Trick : Since $\left| {\frac{3}{2} - 2i} \right| - \left( {\frac{3}{2} - 2i} \right)$
$ = \sqrt {\frac{9}{4} + 4} - \frac{3}{2} + 2i = \frac{5}{2} - \frac{3}{2} + 2i = 1 + 2i$
View full question & answer→MCQ 361 Mark
If $|{z_1}| = |{z_2}| = .......... = |{z_n}| = 1,$ then the value of $|{z_1} + {z_2} + {z_3} + ............. + {z_n}|$=
AnswerCorrect option: C. $\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ......... + \frac{1}{{{z_n}}}} \right|$
c
(c) We have $|{z_k}| = 1,k = 1,\,\,2,....n$
==> $|{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{1}{{{z_k}}}$
Therefore $|{z_1} + {z_2} + .... + {z_n}| = |\overline {{z_1} + {z_2} + .... + {z_n}} |$
$(\because \,\,\,|z|\, = \,|\,\overline z |)$
$ = |{\overline z _1} + \overline {{z_2}} + ..... + {\overline z _n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + .... + \frac{1}{{{z_n}}}} \right|$
Aliter : Let ${z_k} = \cos {\theta _k} + i\sin {\theta _k},\,\,\,k = 1,\,\,2,....n$
So that $|{z_k}| = \sqrt {{{\cos }^2}{\theta _k} + {{\sin }^2}{\theta _k}} = 1$
Then $\frac{1}{{{z_k}}} = {(\cos {\theta _k} + i\sin {\theta _k})^{ - 1}} = (\cos {\theta _k} - i\sin {\theta _k})$
Now, ${z_1} + {z_2} + ..... + {z_n}$
$ = (\cos {\theta _1} + ..... + \cos {\theta _n}) - i(\sin {\theta _1} + ..... + \sin {\theta _n})$
and $\left( {\frac{1}{{{z_1}}}} \right) + \left( {\frac{1}{{{z_2}}}} \right) + ..... + \left( {\frac{1}{{{z_n}}}} \right)$
$ = (\cos {\theta _1} + ..... + \cos {\theta _n}) - i(\sin {\theta _1} + ..... + \sin {\theta _n})$
Hence $|{z_1} + {z_2} + ..... + {z_n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ..... + \frac{1}{{{z_n}}}} \right|$
Since each side is equal to
$\sqrt {{{(\cos {\theta _1} + ..... + \cos {\theta _n})}^2} + {{(\sin {\theta _1} + .... + \sin {\theta _n})}^2}} $
View full question & answer→MCQ 371 Mark
If $z_1, z_2 $ are any two complex numbers, then $|{z_1} + \sqrt {z_1^2 - z_2^2} |$ $ + |{z_1} - \sqrt {z_1^2 - z_2^2} |$ is equal to
AnswerCorrect option: D. $|{z_1} + {z_2}| + |{z_1} - {z_2}|$
d
(d)Trick : Check by putting ${z_1} = 1 + 0i$and ${z_2} = 0 + i$
.
View full question & answer→MCQ 381 Mark
The minimum value of $|2z - 1| + |3z - 2|$is
Answerc
(c) Given expression, $|2z - 1| + |3z - 2|$, minimum value of $|2z - 1|$is $0$ at $z = \frac{1}{2}$. So value of given expression $ = 0 + \frac{1}{2} = \frac{1}{2},$ minimum value of $|3z - 2|$ is $0$, at $z = \frac{2}{3}$. So value of given expression $ = \frac{1}{3} + 0 = \frac{1}{3}$.
So minimum value of given expression is $\frac{1}{3}$.
View full question & answer→MCQ 391 Mark
Let ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$be any complex number, then $\left| {\frac{{{z_1} - {z_2}}}{{1 - {z_1}{{\bar z}_2}}}} \right| = $
Answerb
(b) We have $|{z_1}|\; = 1$ and ${z_2}$be any complex number.
$ \Rightarrow \left| {\;\frac{{{z_1} - {z_2}}}{{1 - {z_1}{{\bar z}_2}}}} \right|\; = \frac{{|{z_1} - {z_2}|}}{{\left| {\;1 - \frac{{{{\bar z}_2}}}{{{{\bar z}_1}}}\;} \right|}}$;$\because \;{z_1}{\bar z_1} = \;|{z_1}{|^2}$
$ = \frac{{|{z_1} - {z_2}|}}{{|{{\bar z}_1} - {{\bar z}_2}|}}|{\bar z_1}|$; Given that $\;|{\bar z_1}|\; = 1$
$ = \frac{{|{z_1} - {z_2}|}}{{|\overline {{z_1} - {z_2}} |}} = \frac{{|{z_1} - {z_2}|}}{{|{z_1} - {z_2}|}} = 1$.
View full question & answer→MCQ 401 Mark
For any two complex numbers ${z_1},{z_2}$we have $|{z_1} + {z_2}{|^2} = $ $|{z_1}{|^2} + |{z_2}{|^2}$ then
- ✓
${\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0$
- B
${\mathop{\rm Im}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0$
- C
${\mathop{\rm Re}\nolimits} ({z_1}{z_2}) = 0$
- D
${\mathop{\rm Im}\nolimits} ({z_1}{z_2}) = 0$
AnswerCorrect option: A. ${\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0$
a
(a) We have $|{z_1} + {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2}$
==> $|{z_1}{|^2} + |{z_2}{|^2} + 2|{z_1}||{z_2}|\cos ({\theta _1} - {\theta _2})$
$ = |{z_1}{|^2} + |{z_2}{|^2}$
Where ${\theta _1} = arg({z_1}),{\theta _2} = arg({z_2})$
==> $\cos ({\theta _1} - {\theta _2}) = 0\,\,\,\, \Rightarrow {\theta _1} - {\theta _2} = \frac{\pi }{2}$
==> $arg\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2} \Rightarrow {\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{{|{z_1}|}}{{|{z_2}|}}\cos \left( {\frac{\pi }{2}} \right) = 0$
Note : Also ${\mathop{\rm Re}\nolimits} \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = 0 \Rightarrow {\mathop{\rm Re}\nolimits} ({z_1}\overline {{z_2}} ) = 0$
==> ${z_1}\overline {{z_2}} $ is purely imaginary.
View full question & answer→MCQ 411 Mark
The sum of amplitude of $z$ and another complex number is $\pi $. The other complex number can be written
- A
$\bar z$
- ✓
$ - \overline z $
- C
$z$
- D
$ - z$
AnswerCorrect option: B. $ - \overline z $
b
(b) We have $z = x + iy$ and let their complex ${z_2}$ and given that $arg\;(z) + {z_2} = \pi $
${z_2} = \pi - arg(z)$;
${z_2} = \pi + \left[ { - {{\tan }^{ - 1}}\frac{y}{x}} \right]$
${z_2} = \pi + [arg\;(\bar z)]$
which lies in second quadrant, i.e. $ - \bar z$.
View full question & answer→MCQ 421 Mark
If ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$ then ${a^2} + {b^2}$ is
Answerc
(c) ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$
Taking modulus and squaring on both sides, we get
${(8 + 1)^{50}} = {3^{98}}({a^2} + {b^2})$
${9^{50}} = {3^{98}}({a^2} + {b^2})$
${3^{100}} = {3^{98}}({a^2} + {b^2})$
==> $({a^2} + {b^2}) = 9$.
View full question & answer→MCQ 431 Mark
If $z$ is a complex number, then the minimum value of $|z| + |z - 1|$ is
Answera
(a)First note that $| - z|\, = \,|z|$ and $|{z_1} + {z_2}|\, \le \,|{z_1}| + |{z_2}|$
Now $|z| + |z - 1|\, = \,|z| + |1 - z|\, \ge \,|z + (1 - z)|\, = \,|1| = 1$
Hence, minimum value of $|z| + |z - 1|$is $1.$
View full question & answer→MCQ 441 Mark
If ${z_1},{z_2}$ and ${z_3},{z_4}$ are two pairs of conjugate complex numbers, then $arg\left( {\frac{{{z_1}}}{{{z_4}}}} \right) + arg\left( {\frac{{{z_2}}}{{{z_3}}}} \right)$ equals
- ✓
$0$
- B
$\frac{\pi }{2}$
- C
$\frac{{3\pi }}{2}$
- D
$\pi $
Answera
(a)We have ${z_2} = {\overline z _1}$and ${z_4} = {\overline z _3}$
Therefore ${z_1}{z_2} = |{z_1}{|^2}$and ${z_3}{z_4} = |{z_3}{|^2}$
Now $arg\left( {\frac{{{z_1}}}{{{z_4}}}} \right) + arg\left( {\frac{{{z_2}}}{{{z_3}}}} \right) = arg\left( {\frac{{{z_1}{z_2}}}{{{z_4}{z_3}}}} \right)$
$ = arg\left( {\frac{{|{z_1}{|^2}}}{{|{z_3}{|^2}}}} \right) = arg\left( {{{\left| {\frac{{{z_1}}}{{{z_3}}}} \right|}^2}} \right)= 0$
(Argument of positive real number is zero).
View full question & answer→MCQ 451 Mark
If complex number $z = x + iy$ is taken such that the amplitude of fraction $\frac{{z - 1}}{{z + 1}}$ is always $\frac{\pi }{4}$, then
AnswerCorrect option: D. ${x^2} + {y^2} - 2y = 1$
d
(d) $\frac{{z - 1}}{{z + 1}} = \frac{{(x + iy) - 1}}{{(x + iy) + 1}} = \frac{{(x - 1) + iy}}{{(x + 1) + iy}}$
$ = \,\frac{{\{ (x - 1) + iy\} \,\{ (x + 1) - iy\} }}{{\{ (x + 1) + iy\} \,\{ (x + 1) - iy\} }}$
$ = \frac{{\{ ({x^2} - 1) + {y^2}\} + i\{ y(x + 1) - y(x - 1)\} }}{{{{(x + 1)}^2} + {y^2}}}$
$ = \left\{ {\frac{{({x^2} - 1) + {y^2}}}{{{{(x + 1)}^2} + {y^2}}}} \right\}$$ + i\,\left\{ {\frac{{2y}}{{{{(x + 1)}^2} + {y^2}}}} \right\}$
$\therefore \,\,$$amp\left( {\frac{{z - 1}}{{z + 1}}} \right)$$ = {\tan ^{ - 1}}\left\{ {\frac{{2y}}{{{{(x + 1)}^2} + {y^2}}} \div \frac{{({x^2} - 1) + {y^2}}}{{{{(x + 1)}^2} + {y^2}}}} \right\}$
==> $\frac{\pi }{4} = {\tan ^{ - 1}}\left\{ {\frac{{2y}}{{{x^2} + {y^2} - 1}}} \right\}$==> $\tan \frac{\pi }{4} = \frac{{2y}}{{{x^2} + {y^2} - 1}}$
==> $1 = \frac{{2y}}{{{x^2} + {y^2} - 1}}$
==> ${x^2} + {y^2} - 1 = 2y$
==> ${x^2} + {y^2} - 2y = 1$.
View full question & answer→MCQ 461 Mark
$\sqrt { - 8 - 6i} = $
- A
$1 \pm 3i$
- ✓
$ \pm (1 - 3i)$
- C
$ \pm (1 + 3i)$
- D
$ \pm (3 - i)$
AnswerCorrect option: B. $ \pm (1 - 3i)$
b
(b) Given that $\sqrt { - 8 - 6i} = x + iy = z$
==> $ - 8 - 6i = {(x + iy)^2}$
$\therefore {x^2} - {y^2} = - 8$ .....$(i) $ and $2xy = - 6$ .....$(ii)$
Now ${x^2} + {y^2} = \sqrt {64 + 36} = \pm 10$ .....$(iii)$
From $(i) $ and $(iii)$, we get $x = \pm 1$and $y = \pm 3$
Hence $z = \pm (1 - 3i)$
Trick : Since ${\{ \pm (1 - 3i)\} ^2} = - 8 - 6i$
View full question & answer→MCQ 471 Mark
If ${( - 7 - 24i)^{1/2}} = x - iy,$ then ${x^2} + {y^2} = $
Answerb
(b) $\sqrt { - 7 - 24i} = x - iy$
Squaring both sides, $ - 7 - 24i = {x^2} - {y^2} - i(2xy)$
Equating real and imaginary parts, we get
${x^2} - {y^2} = - 7$and $2xy = 24$
$\therefore \,\,\,{x^2} + {y^2} = \sqrt {49 + 576} = \sqrt {625} = 25$
View full question & answer→MCQ 481 Mark
If $\sqrt {x + iy} = \pm (a + ib),$ then $\sqrt { - x - iy} $ is equal to
- A
$ \pm (b + ia)$
- B
$ \pm (a - ib)$
- ✓
$ \pm (b - ia)$
- D
AnswerCorrect option: C. $ \pm (b - ia)$
c
(c) $\sqrt {x + iy} = \pm (a + bi)$
==> $x + iy = {a^2} - {b^2} + 2iab$ ==> $x = {a^2} - {b^2},$$y = 2ab$
$\sqrt { - x - iy} = \sqrt { - ({a^2} - {b^2}) - 2iab} = \sqrt {{b^2} - {a^2} - 2iab} $
$ = \sqrt {{{(b - ia)}^2}} = \pm (b - ia)$.
View full question & answer→MCQ 491 Mark
The number of non-zero integral solutions of the equation $|1 - i{|^x} = {2^x}$ is
Answerd
(d) Since $1 - i = \sqrt 2 \left\{ {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right\},|1 - i| = \sqrt 2 $
$\therefore $$|1 - i{|^x} = {2^x}$==>${(\sqrt 2 )^x} = {2^x}$==> ${2^{x/2}} = {2^x}$
==> $\frac{x}{2} = x$==> $x = 0$
Therefore, the number of non-zero integral solutions is nil or zero.
View full question & answer→MCQ 501 Mark
$\frac{{1 + 7i}}{{{{(2 - i)}^2}}} = $
- ✓
$\sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$
- B
$\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)$
- C
$\left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$
- D
AnswerCorrect option: A. $\sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$
a
(a) $\frac{{1 + 7i}}{{{{(2 - i)}^2}}} = \frac{{(1 + 7i)}}{{(3 - 4i)}}\frac{{(3 + 4i)}}{{(3 + 4i)}} = \frac{{ - 25 + 25i}}{{25}} = - 1 + i$
Let $z = x + iy = - 1 + i$
$\therefore r\cos \theta = - 1$and $r\sin \theta =1$ $\therefore \theta = \frac{{3\pi }}{4}$and $r = \sqrt 2 $
Thus $\frac{{1 + 7i}}{{{{(2 - i)}^2}}} = \sqrt 2 \left[ {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right]$
Aliter : $\left| {\frac{{1 + 7i}}{{{{(2 - i)}^2}}}} \right| = \left| {\frac{{1 + 7i}}{{3 - 4i}}} \right| = \sqrt 2 $
and $arg\left( {\frac{{1 + 7i}}{{3 - 4i}}} \right) = {\tan ^{ - 1}}7 - {\tan ^{ - 1}}\left( { - \frac{4}{3}} \right)$
$ = {\tan ^{ - 1}}7 + {\tan ^{ - 1}}\frac{4}{3} = \frac{{3\pi }}{4}$
$\therefore \frac{{1 + 7i}}{{{{(2 - i)}^2}}}$$ = \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$
View full question & answer→