MCQ 11 Mark
If z = x + iy and |z – zi| = 1, then
MCQ 21 Mark
If $-1+\sqrt{ } 3 i = re ^{ i \theta}$, then $\theta=$
- A
$-\frac{2 \pi}{3}$
- B
$\frac{\pi}{3}$
- C
$-\frac{\pi}{3}$
- ✓
$\frac{2 \pi}{3}$
AnswerCorrect option: D. $\frac{2 \pi}{3}$
$\frac{2 \pi}{3}$
View full question & answer→MCQ 31 Mark
If $\arg ( z )=\theta$, then $\arg \overline{( z )}=$
View full question & answer→MCQ 41 Mark
The modulus and argument of $(1+ i \sqrt{ } 3)^8$ are respectively
AnswerCorrect option: C. 256 and $\frac{2 \pi}{3}$
256 and $\frac{2 \pi}{3}$
View full question & answer→MCQ 51 Mark
If $\omega(\neq 1)$ is a cube root of unity and $(1+\omega)^7=A+B \omega$, then $A$ and $B$ are respectively the numbers
View full question & answer→MCQ 61 Mark
If $z=r(\cos \theta+i \sin \theta)$, then the value of $\frac{z}{\bar{z}}+\frac{\bar{z}}{z}$ is
View full question & answer→MCQ 71 Mark
If $\omega$ is a complex cube root of unity, then the value of $\omega^{99}+\omega^{100}+\omega^{101}$ is:
Answer0
$\omega^{99}+\omega^{100}+\omega^{101} $
$ =\omega^{99}\left(1+\omega+\omega^2\right) $
$ =\omega^{99}(0) \\ =0$
View full question & answer→MCQ 81 Mark
Answer -3√2
Hint: √-3 √-6 = (√3 i) (√6 i) = 3√2 (-1) = -3√2
View full question & answer→MCQ 91 Mark
The value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{554}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}$ is equal to:
Answer$.-1$
$\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}} $ $=\frac{i^{53}\left[i^8+i^6+i^4+i^2+1\right]}{i^{574}\left[i^8+i^6+i^4+i^2+1\right]}$
$=i^{10}$
$=\left(i^2\right)^5$
$=(-1)^5=-1$
View full question & answer→MCQ 101 Mark
If $n$ is an odd positive integer, then the value of $1+(i)^{2 n}+(i)^{4 n}+(i)^{6 n}$ is:
Answer0
$1+\left(i^2\right)^n+\left(i^4\right)^n+\left(i^2\right)^{3 n} $
$ =1-1+1-1 \ldots . .(n \text { odd positive integer })$
View full question & answer→MCQ 112 Marks
If $Z_1=4 i^{40}-5 i^{35}+6 i^{17}+2, Z_2=-1+i$, where $i=\sqrt{-1}$, then $\left|Z_1+Z_2\right|=$
Answer
(b) : Given, $z_1=4 i^{40}-5 i^{35}+6 i^{17}+2$
and $z_2=-1+i$
$
\begin{aligned}
& \Rightarrow z_1=4 i^{4 \times 10}-5 i^{4 \times 8+3}+6 i^{4 \times 4+1}+2 \\
& =4+5 i+6 i+2 \quad \quad\left[\because i^{4n}=1, \forall n \in N \right] \\
& =6+11 i
\end{aligned}
$
Now, $z_1+z_2=6+11 i-1+i$
$
=5+12 i
$
$
\therefore\left|z_1+z_2\right|=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13
$
View full question & answer→MCQ 122 Marks
Let $z \in C$ with $\operatorname{Im}(z)=10$ and its satisfies $\frac{2 z-n}{2 z+n}=2 i-1, i=\sqrt{-1}$ for some natural number $n$, then
- A
$n=20$ and $\operatorname{Re}(z)=-10$
- ✓
$n=40$ and $\operatorname{Re}(z)=-10$
- C
$n=40$ and $\operatorname{Re}(z)=10$
- D
$n=20$ and $\operatorname{Re}(z)=10$
AnswerCorrect option: B. $n=40$ and $\operatorname{Re}(z)=-10$
(b) : Let $\operatorname{Re}(z)=x$ and $\operatorname{Im}(z)=10$ (given)
$\therefore z=x+10 i$
Now, $\frac{2 z-n}{2 z+n}=2 i-1$
$
\begin{aligned}
& \Rightarrow 2(x+10 i)-n=(2 i-1)(2(x+10 i)+n) \\
& \Rightarrow 2 x+20 i-n=(2 i-1)((2 x+n)+20 i) \\
& \Rightarrow 2 x+20 i-n=(4 x+2 n) i-40-(2 x+n)-20 i \\
& \Rightarrow(2 x-n)+20 i=-(2 x+n+40)+(4 x+2-20) i
\end{aligned}
$
On comparing the real and imaginary part, we get
$
\begin{aligned}
& 2 x-n=-2 x-n-40 \text { and } 4 x+2 n-20=20 \\
\Rightarrow & 4 x=-40 \Rightarrow 4 x+2 n=40 \\
\Rightarrow & x=-10 \Rightarrow-40+2 n=40 \\
\Rightarrow & 2 n=80 \Rightarrow n=40 \\
\therefore & n=40 \text { and } \operatorname{Re}(z)=-10
\end{aligned}
$
View full question & answer→MCQ 132 Marks
If $a>0$ and $z=\frac{(1+i)^2}{a-i}, i=\sqrt{-1}$, has magnitude $\frac{2}{\sqrt{5}}$, then $\bar{z}$ is
- ✓
$-\frac{2}{5}-\frac{4}{5} i$
- B
$-\frac{2}{5}+\frac{4}{5} i$
- C
$\frac{2}{5}-\frac{4}{5} i$
- D
$\frac{2}{5}+\frac{4}{5} i$
AnswerCorrect option: A. $-\frac{2}{5}-\frac{4}{5} i$
(a) : $z=\frac{(1+i)^2}{a-i}$
On rationalising the denominator, we get
$
z=\frac{(1+i)^2}{a-i} \times \frac{a+i}{a+i}=\frac{(1-1+2 i)(a+i)}{a^2+1}=\frac{2 a i-2}{a^2+1}
$
$\therefore z=\frac{-2}{a^2+1}+\frac{2 a}{a^2+1} i \Rightarrow|z|=\sqrt{\frac{4}{\left(a^2+1\right)^2}+\frac{4 a^2}{\left(a^2+1\right)^2}}$
$=\sqrt{\frac{4\left(a^2+1\right)}{\left(a^2+1\right)^2}}=\frac{2}{\sqrt{a^2+1}}=\frac{2}{\sqrt{5}}$ (Given)
$\Rightarrow a^2+1=5 \Rightarrow a^2+1=2^2+1 \Rightarrow a=2$
$\therefore z=\frac{-2}{5}+\frac{4}{5} i \Rightarrow \bar{z}=\frac{-2}{5}-\frac{4}{5} i$
View full question & answer→MCQ 142 Marks
If the Cartesian co-ordinates of a point are $\left(\frac{-5 \sqrt{3}}{2}, \frac{5}{2}\right)$, then its polar co-ordinates are
- ✓
$\left(5, \frac{5 \pi}{6}\right)$
- B
$\left(5, \frac{2 \pi}{3}\right)$
- C
$\left(5, \frac{11 \pi}{18}\right)$
- D
$\left(5, \frac{13 \pi}{18}\right)$
AnswerCorrect option: A. $\left(5, \frac{5 \pi}{6}\right)$
(a) : If $(x, y)$ are cartesian coordinates then polar coordinates can be written as $(r, \theta)$, where
$
r=\sqrt{x^2+y^2}, \theta=\tan ^{-1}\left(\frac{y}{x}\right)
$
Given, $x=\frac{-5 \sqrt{3}}{2}, y=\frac{5}{2}$
So, $r=\sqrt{\left(\frac{-5 \sqrt{3}}{2}\right)^2+\left(\frac{5}{2}\right)^2}=5, \theta=\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=\frac{5 \pi}{6}$
Hence, polar coordinates are $\left(5, \frac{5 \pi}{6}\right)$
Complex Numbers
View full question & answer→MCQ 152 Marks
The Cartesian form of complex number $z=4\left(\cos 300^{\circ}+i \sin 300^{\circ}\right)$, where $i=\sqrt{-1}$ is
- ✓
$2-2 \sqrt{3} i$
- B
$2+2 \sqrt{3} i$
- C
$1-\sqrt{3} i$
- D
$1+\sqrt{3} i$
AnswerCorrect option: A. $2-2 \sqrt{3} i$
(a) : Given, $z=4\left(\cos 300^{\circ}+i \sin 300^{\circ}\right)$
$=4\left[\cos \left(270^{\circ}+30^{\circ}\right)+i \sin \left(270^{\circ}+30^{\circ}\right)\right]$
$=4\left[\sin 30^{\circ}+i\left(-\cos 30^{\circ}\right)\right]$
$=4\left[\frac{1}{2}-\frac{\sqrt{3}}{2} i\right]=2-2 \sqrt{3} i$
View full question & answer→MCQ 162 Marks
If $(x-i y)(3+5 i)$ is the conjugate of $-6-24 i$ (where $x, y \in R$ and $i=\sqrt{-1}$ ), then the values of $x$ and $y$ are respectively.
AnswerCorrect option: B. $3,-3$
(b) : Let $z_1=(x-i y)(3+5 i)$ and $z_2=-6-24 i$
Given, $\bar{z}_2=z_1$
$
\begin{aligned}
& \Rightarrow-6+24 i=3 x+5 x i-3 y i+5 y \\
& \Rightarrow-6+24 i=(3 x+5 y)+(5 x-3 y) i \\
& \Rightarrow 3 x+5 y=-6 \text { and } 5 x-3 y=24
\end{aligned}
$
Solving these two equations, we get $x=3, y=-3$
View full question & answer→MCQ 172 Marks
In the Argand plane, the distinct roots of $1+z+z^3+z^4=0$ ( $z$ is a complex number) represent vertices of
Answer(B)
$1+z+z^3+z^4=0$
$\Rightarrow(1+z)\left(1+z^3\right)=0$
$\Rightarrow z =-1,-1,-\omega,-\omega^2$, where $\omega$ is a cube root of unity.
$\therefore \quad$ the distinct roots are $(-1,0),\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$.
Distance between each of them is $\sqrt{3}$. So, they form an equilateral triangle.
View full question & answer→MCQ 182 Marks
If $\alpha, \beta, \gamma$ are the cube roots of $p ( p <0)$, then for any $x, y$ and $z, \frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}=$
- ✓
$\frac{1}{2}(-1- i \sqrt{3})$
- B
$\frac{1}{2}(1+ i \sqrt{3})$
- C
$\frac{1}{2}(1- i \sqrt{3})$
- D
$\frac{1}{2}(-1+i \sqrt{3})$
AnswerCorrect option: A. $\frac{1}{2}(-1- i \sqrt{3})$
(A)
Since $p <0$. Let $p =- q$, where q is positive.
Therefore $p^{\frac{1}{3}}=-q^{\frac{1}{3}}(1)^{\frac{1}{3}}$
Hence $\alpha=-q^{\frac{1}{3}}, \beta=-q^{\frac{1}{3}} \omega$ and $\gamma=-q^{\frac{1}{3}} \omega^2$
The given expression $=\frac{x+y \omega+z \omega^2}{x \omega+y \omega^2+z}$
$=\frac{1}{\omega} \cdot \frac{x \omega+y \omega^2+z}{x \omega+y \omega^2+z}$
$=\frac{1}{\omega} \cdot 1=\frac{1}{\omega} \cdot \omega^3$
$=\omega^2$
$=\frac{1}{2}(-1- i \sqrt{3})$
View full question & answer→MCQ 192 Marks
If $\omega_n=\cos \left(\frac{2 \pi}{n}\right)+i \sin \left(\frac{2 \pi}{n}\right), i^2=-1$, then $\left(x+y \omega_3+z \omega_3{ }^2\right)\left(x+y \omega_3{ }^2+z \omega_3\right)$ is equal to
AnswerCorrect option: C. $x^2+y^2+z^2-y z-z x-x y$
(C)
$\omega_n=\cos \left(\frac{2 \pi}{n}\right)+i \sin \left(\frac{2 \pi}{n}\right)$
$\Rightarrow \omega_3=\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}=-\frac{1}{2}+\frac{i \sqrt{3}}{2}=\omega$
and $\omega_3^2-\left(\cos \frac{2 \pi}{3}+ \text { i } \sin \frac{2 \pi}{3}\right)^2$
$=\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)=-\frac{1}{2}-\frac{i \sqrt{3}}{2}=\omega^2$
$\therefore \left(x+y \omega_3+z \omega_3^2\right)\left(x+y \omega_3^2+z \omega_3\right)$
$=\left(x+y \omega+z \omega^2\right)\left(x+y \omega^2+z \omega\right)$
$=x^2+y^2+z^2-x y-y z-z x$
View full question & answer→MCQ 202 Marks
If $z^2+z+1=0$, where $z$ is a complex number, then the value of $\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2+\ldots+\left(z^6+\frac{1}{z^6}\right)^2$ is
Answer(D)
$z^2+z+1=0 \Rightarrow z=\omega$ or $\omega^2$...[Using $x^2 \pm x+1=(x \mp \omega)\left(x \mp \omega^2\right)$
Let $z =\omega$
$\therefore\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\ldots .+\left(z^6+\frac{1}{z^6}\right)^2$
$=\left(\omega+\frac{1}{\omega}\right)^2+\left(\omega^2+\frac{1}{\omega^2}\right)^2+\ldots .+\left(\omega^6+\frac{1}{\omega^6}\right)^2$
$=1+1+4+1+1+4=12$
View full question & answer→MCQ 212 Marks
If $z + z ^{-1}=1$, then $z ^{100}+ z ^{-100}$ is equal to
Answer(D)
$z+z^{-1}=1 \Rightarrow z^2-z+1=0$
$\Rightarrow z=-\omega \text { or }-\omega^2$…[Using $x^2 \pm x+1=(x \mp \omega)\left(x \mp \omega^2\right)$
For $z=-\omega, z^{100}+z^{-100}=(-\omega)^{100}+(-\omega)^{-100}$
$=\omega+\frac{1}{\omega}$
$=\frac{\omega^2+1}{\omega}=-1$
For $z=-\omega^2, z^{100}+z^{-100}=\left(-\omega^2\right)^{100}+\left(-\omega^2\right)^{-100}$
$=\omega^{200}+\frac{1}{\omega^{200}}$
$=\omega^2+\frac{1}{\omega^2}$
$=\omega^2+\omega=-1$
View full question & answer→MCQ 222 Marks
$\frac{(-1+i \sqrt{3})^{15}}{(1-i)^{20}}+\frac{(-1-i \sqrt{3})^{15}}{(1+i)^{20}}$ is equal to
- ✓
$-64$
- B
$-32$
- C
$-16$
- D
$\frac{1}{16}$
Answer(A)
$2^{15}\left[\frac{\left(-\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{15}}{(1- i )^{20}}+\frac{\left(\frac{-1}{2}-\frac{ i \sqrt{3}}{2}\right)^{15}}{(1+ i )^{20}}\right]$
$=2^{15}\left[\frac{\omega^{15}}{(1- i )^{20}}+\frac{\omega^{30}}{(1+ i )^{20}}\right]$
$=2^{15}\left[\frac{1}{(1-i)^{20}}+\frac{1}{(1+i)^{20}}\right]$
$=2^{15}\left[\frac{(1+i)^{20}+(1-i)^{20}}{\left(1-i^2\right)^{20}}\right]$
$=\frac{2^{15}}{2^{20}}\left[(1+i)^{20}+(1-i)^{20}\right]$
$=\frac{1}{2^5}\left[\left\{(1+i)^2\right\}^{10}+\left\{(1-i)^2\right\}^{10}\right]$
$=\frac{1}{2^5}\left[(2 i )^{10}+(-2 i )^{10}\right]=\frac{2^{11} \cdot 1^{10}}{2^5}$
$=-2^6 \quad \ldots\left[\because i ^{10}=\left( i ^4\right)^2 \cdot i ^2= i ^2=-1\right]$
$=-64$
View full question & answer→MCQ 232 Marks
The value of $\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64}$ is
Answer(B)
$1+\sqrt{3} i=2 \omega^2$
$1-\sqrt{3} i=-2 \omega$
$\omega^{64}+\frac{1}{\omega^{64}}=\omega+\omega^2=-1$
View full question & answer→MCQ 242 Marks
$\left(\frac{\sqrt{3}+ i }{2}\right)^6+\left(\frac{ i -\sqrt{3}}{2}\right)^6$ is equal to
Answer(A)
$\left(\frac{\sqrt{3}+ i }{2}\right)^6+\left(\frac{ i -\sqrt{3}}{2}\right)^6$
$=\left(\frac{-1+\sqrt{3} i }{2 i }\right)^6+\left(\frac{-1-\sqrt{3} i }{2 i }\right)^6$
$=\frac{1}{i^6}\left[(\omega)^6+\left(\omega^2\right)^6\right]$
$\ldots\left[\because \omega=\frac{-1+\sqrt{3} i }{2}, \omega^2=\frac{-1-\sqrt{3} i }{2}\right]$
$=-\left[\left(\omega^3\right)^2+\left(\omega^3\right)^4\right]=-(1+1)=-2$
View full question & answer→MCQ 252 Marks
If $z=\frac{\sqrt{3}+i}{2}$, then $z^{69}$ is equal to
Answer(A)
$z^{69}=\left(\frac{\sqrt{3}+i}{2}\right)^{69}=\left[\frac{1}{i}\left(\frac{-1+\sqrt{3} i}{2}\right)\right]^{69}$
$=\left(\frac{\omega}{i}\right)^{69}$
$=\frac{\omega^{69}}{\left(i^4\right)^{17} i}=\frac{1}{i}=-i$
View full question & answer→MCQ 262 Marks
If $i =\sqrt{-1}$, then $4+5\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{365}=$
- A
$1-i \sqrt{3}$
- B
$-1+i \sqrt{3}$
- ✓
$i \sqrt{3}$
- D
$- i \sqrt{3}$
AnswerCorrect option: C. $i \sqrt{3}$
(C)
$4+5\left(-\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{ i \sqrt{3}}{2}\right)^{365}$
$=4+5 \omega^{334}+3 \omega^{365}$
$=4+5 \omega+3 \omega^2$
$=4+5\left(-\frac{1}{2}+ i \frac{\sqrt{3}}{2}\right)+3\left(-\frac{1}{2}- i \frac{\sqrt{3}}{2}\right)$
$= i \sqrt{3}$
View full question & answer→MCQ 272 Marks
Value of $\left(\frac{-1+\sqrt{-3}}{2}\right)^{40}+\left(\frac{-1-\sqrt{-3}}{2}\right)^{40}$ is
Answer(D)
Using $\left(\frac{-1+ i \sqrt{3}}{2}\right)^{ n }+\left(\frac{-1- i \sqrt{3}}{2}\right)^{ n }=-1$
i.e., $\omega^{ n }+\omega^{2 n }=-1$ if n is $a + ve$ integer other than multiple of 3. , we get
$\left(\frac{-1+\sqrt{-3}}{2}\right)^{40}+\left(\frac{-1-\sqrt{-3}}{2}\right)^{40}=-1$
View full question & answer→MCQ 282 Marks
If $\left(\frac{1+ i \sqrt{3}}{1- i \sqrt{3}}\right)^{ n }$ is an integer, then n is
Answer(C)
$\frac{1+i \sqrt{3}}{1-i \sqrt{3}}=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)\left(\frac{1+i \sqrt{3}}{1+i \sqrt{3}}\right)$
$=\frac{-2+ i 2 \sqrt{3}}{4}$
$=\frac{-1+i \sqrt{3}}{2}=\omega$
$\therefore \quad\left(\frac{1+ i \sqrt{3}}{1- i \sqrt{3}}\right)^{ n }=\omega^{ n }=\omega^3=1 \Rightarrow n =3$
View full question & answer→MCQ 292 Marks
If $z=e^{i 4 \pi / 3}$, then $\left(z^{192}+z^{194}\right)^3$ is equal to
Answer(B)
$Z = e ^{ i \left(\frac{4 \pi}{3}\right)}$
$=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}$
$=-\frac{1}{2}-\frac{\sqrt{3}}{2} i =\omega^2$
$\therefore \quad\left( z ^{192}+ z ^{194}\right)^3=\left[\left(\omega^2\right)^{192}+\left(\omega^2\right)^{194}\right]^3$
$=\left[\left(\omega^3\right)^{128}+\left(\omega^3\right)^{129} \cdot \omega\right]^3$
$=(1+\omega)^3$
$=\left(-\omega^2\right)^3 \quad \ldots\left[\because 1+\omega+\omega^2=0\right]$
$=-\omega^6$
$=-1$
View full question & answer→MCQ 302 Marks
If $2 x=-1+\sqrt{3} i$, then the value of $\left(1-x^2+x\right)^6-\left(1-x+x^2\right)^6$ is
Answer(D)
Given, $x=\frac{-1+\sqrt{3} i }{2} \Rightarrow x=\omega$
$\therefore \quad\left(1-x^2+x\right)^6-\left(1-x+x^2\right)^6$
$=\left(1-\omega^2+\omega\right)^6-\left(1-\omega+\omega^2\right)^6$
$=\left((1+\omega)-\omega^2\right)^6-\left(\left(1+\omega^2\right)-\omega\right)^6$
$=\left(-\omega^2-\omega^2\right)^6-(-\omega-\omega)^6$
$=\left(-2 \omega^2\right)^6-(-2 \omega)^6$
$=64 \omega^{12}-64 \omega^6$
$=64\left(\omega^3\right)^4-64\left(\omega^3\right)^2=0 \quad \ldots .\left[\because \omega^3=1\right]$
View full question & answer→MCQ 312 Marks
If $\omega$ is an imaginary cube root of unity, then the value of $\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]$ is
- A
$-\sqrt{3} / 2$
- B
$-1 / \sqrt{2}$
- ✓
$1 / \sqrt{2}$
- D
$\sqrt{3} / 2$
AnswerCorrect option: C. $1 / \sqrt{2}$
(C)
$\sin \left[\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right]=\sin \left[\left(\omega+\omega^2\right) \pi-\frac{\pi}{4}\right]$
$=\sin \left(-\pi-\frac{\pi}{4}\right)=-\sin \left(\pi+\frac{\pi}{4}\right)=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 322 Marks
If $a + b + c =0$ and $1, \omega, \omega^2$ are three cube roots of unity, then $\left(a+b \omega+c \omega^2\right)^3+\left(a+b \omega^2+c \omega\right)^3$ is equal to
- ✓
- B
$-3 a b c$
- C
- D
$-27 a b c$
Answer(A)
Put $a =1, b=1, c =-2, \quad \because a + b + c =0$
$\therefore \quad\left(1+\omega-2 \omega^2\right)^3+\left(1+\omega^2-2 \omega\right)^3$
$=\left(-3 \omega^2\right)^3+(-3 \omega)^3$
$=-27-27=-54$
Also, option (A) gives the value -54
i.e., $27 \times 1 \times 1 \times(-2)=-54$
View full question & answer→MCQ 332 Marks
If $\omega$ is a complex cube root of unity, then $(x+y)^3+\left(x \omega+y \omega^2\right)^3+\left(x \omega^2+y \omega\right)^3=$
- ✓
$3\left(x^3+y^3\right)$
- B
$3\left(x^3-y^3\right)$
- C
$4\left(x^3+y^3\right)$
- D
$4\left(x^3-y^3\right)$
AnswerCorrect option: A. $3\left(x^3+y^3\right)$
(A)
After solving, we get
$3 x^3+3 y^3+3 x^2 y\left(1+\omega+\omega^2\right)+3 x y^2\left(1+\omega+\omega^2\right)$
$=3\left(x^3+y^3\right)+3.0+3.0$
$=3\left(x^3+y^3\right)$
View full question & answer→MCQ 342 Marks
If $1, \omega$ and $\omega^2$ are the cube roots of unity, then $(a+b+c)\left(a+b \omega+c \omega^2\right)\left(a+b \omega^2+c \omega\right)=$
- A
$a^3+b^3+c^3$
- ✓
$a^3+b^3+c^3-3 a b c$
- C
$(a+b+c)^3-3 a b c$
- D
$a^3+b^3+c^3+3 a b c$
AnswerCorrect option: B. $a^3+b^3+c^3-3 a b c$
(B)
$\left(a+b \omega+c \omega^2\right)\left(a+b \omega^2+c \omega\right)$
$=a^2+a b \omega^2+a c \omega+a b \omega+b^2 \omega^3+b c \omega^2$ $+a c \omega^2+b c \omega^4+c^2 \omega^3$
$=a^2+b^2 \omega^3+c^2 \omega^3+(a b+b c+a c) \omega^2$ $+(a b+b c+a c) \omega$
$=a^2+b^2+c^2+(a b+b c+a c)\left(\omega^2+\omega\right)$
$=a^2+b^2+c^2-(a b+b c+a c)$
$\therefore \quad( a + b + c )\left( a + b \omega+ c \omega^2\right)\left( a + b \omega^2+ c \omega\right)$
$=a^3+b^3+c^3-3 a b c$
View full question & answer→MCQ 352 Marks
If $\omega$ is a non real cube root of unity, then $(a+b)(a+b \omega)\left(a+b \omega^2\right)$ is
- ✓
$a^3+b^3$
- B
$a^3-b^3$
- C
$a^2+b^2$
- D
$a^2-b^2$
AnswerCorrect option: A. $a^3+b^3$
(A)
$(a+b)(a+b \omega)\left(a+b \omega^2\right)$
$=(a+b)\left(a^2+a b\left(\omega+\omega^2\right)+b^2 \omega^3\right)$
$=( a + b )\left( a ^2- ab + b ^2\right)= a ^3+ b ^3$
View full question & answer→MCQ 362 Marks
Let $\omega$ be the imaginary root of $x^n=1$, then $(5-\omega)\left(5-\omega^2\right) \ldots\left(5-\omega^{ n -1}\right)$ is equal to
- A
- B
$\frac{5^n+1}{4}$
- C
$4^{n-1}$
- ✓
$\frac{5^n-1}{4}$
AnswerCorrect option: D. $\frac{5^n-1}{4}$
(D)
If $\omega$ is an imaginary (non-real) $n ^{\text {th }}$ root of unity, then all the $n ^{\text {th }}$ roots are
$1, \omega, \omega^2, \ldots, \omega^{ n -1}$
$\therefore \quad x^{ n }-1=(x-1)(x-\omega)\left(x-\omega^2\right) \ldots\left(x-\omega^{ n -1}\right)$
Substituting $x=5$, we get
$5^n-1=(5-1)(5-\omega)\left(5-\omega^2\right)\left(5-\omega^2\right)$ $\ldots\left(5-\omega^{n-1}\right)$
$\therefore \quad \frac{5^n-1}{4}=(5-\omega)\left(5-\omega^2\right) \ldots . .\left(5-\omega^{n-1}\right)$
View full question & answer→MCQ 372 Marks
If $\omega$ is a complex cube root of unity, then $(2-\omega)\left(2-\omega^2\right)\left(2-\omega^{10}\right)\left(2-\omega^{11}\right)$ is
Answer(C)
$(2-\omega)\left(2-\omega^2\right)(2-\omega)\left(2-\omega^2\right)$
$=(2-\omega)^2\left(2-\omega^2\right)^2$
$=\left[(2-\omega)\left(2-\omega^2\right)\right]^2$
$=\left[4-2\left(\omega+\omega^2\right)+\omega^3\right]^2$
$=[4+2+1]^2$
$=49$
View full question & answer→MCQ 382 Marks
If $\alpha$ and $\beta$ are complex cube roots of unity, then $(1-\alpha)(1-\beta)\left(1-\alpha^2\right)\left(1-\beta^2\right)=$
Answer(C)
$(1-\alpha)(1-\beta)\left(1-\alpha^2\right)\left(1-\beta^2\right)$
$=(1-\alpha)\left(1-\alpha^2\right)\left(1-\alpha^2\right)(1-\alpha)$
$=(1-\alpha)^2\left(1-\alpha^2\right)^2=(-\alpha-2 \alpha)\left(1-2 \alpha^2+\alpha\right)$
$=(-3 \alpha)\left(-\alpha^2-2 \alpha^2\right)=(-3 \alpha)\left(-3 \alpha^2\right)=9$
View full question & answer→MCQ 392 Marks
If $\alpha$ and $\beta$ are imaginary cube roots of unity, then the value of $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$ is
Answer(C)
Since $\alpha$ and $\beta$ are complex roots of unity,
we may write $\alpha=\omega, \beta=\omega^2$
$\therefore \quad \alpha^4+\beta^{28}+\frac{1}{\alpha \beta}=\omega^4+\left(\omega^2\right)^{28}+\frac{1}{\omega \cdot \omega^2}$
$=\omega+\omega^{56}+1$
$=\omega+\omega^2+1$
$=0$
View full question & answer→MCQ 402 Marks
If $\alpha$ and $\beta$ are the roots of the equation $x^2-x+1=0$, then $\alpha^{2009}+\beta^{2009}=$
Answer(C)
Roots of the equation $x^2-x+1=0$ are
$\alpha=-\omega, \beta=-\omega^2$ ...[Using $x^2 \pm x+1=(x \mp \omega)\left(x \mp \omega^2\right)$]
$\therefore \quad \alpha^{2009}+\beta^{2009}=(-\omega)^{2009}+\left(-\omega^2\right)^{2009}$
$=-\left(\omega^2+\omega\right)=1$
View full question & answer→MCQ 412 Marks
If $\alpha$ is a complex cube root of unity such that $\alpha^2+\alpha+1=0$, then $\alpha^{31}$ is
- ✓
$\alpha$
- B
$\alpha^2$
- C
$0$
- D
AnswerCorrect option: A. $\alpha$
(A)
$\alpha^2+\alpha+1=0$
$\therefore \quad(\alpha-1)\left(\alpha^2+\alpha+1\right)=0$
$\therefore \quad \alpha^3-1=0, \alpha \neq 1$
$\Rightarrow \alpha^3=1$
and consequently $\alpha^{31}=\left(\alpha^3\right)^{10} \cdot \alpha=1^{10} \alpha=\alpha$
View full question & answer→MCQ 422 Marks
If $1, \omega, \omega^2$ are the cube roots of unity and if $\alpha=\omega+2 \omega^2-3$ then $\alpha^3+12 \alpha^2+48 \alpha+3=$
Answer(D)
$\alpha=\omega+2 \omega^2-3$
$\Rightarrow \alpha=-4+\omega^2$
$\Rightarrow \alpha+4=\omega^2$
$\Rightarrow(\alpha+4)^3=\left(\omega^2\right)^3$
$\Rightarrow \alpha^3+12 \alpha^2+48 \alpha+3=-60$
View full question & answer→MCQ 432 Marks
If the cube roots of unity are $1, \omega, \omega^2$, then the roots of the equation $(x-2)^3+27=0$ are
- A
$-1,-1,-1$
- B
$-1,-\omega,-\omega^2$
- C
$-1,2+3 \omega, 2+3 \omega^2$
- ✓
$-1,2-3 \omega, 2-3 \omega^2$
AnswerCorrect option: D. $-1,2-3 \omega, 2-3 \omega^2$
(D)
Here, $1^{\frac{1}{3}}=1, \omega, \omega^2$
$\therefore \quad$ For the equation $(x-2)^3+27=0$
$\Rightarrow(x-2)^3=-27=-3^3$
$\Rightarrow x-2=-3(1)^{\frac{1}{3}}=-3\left(1, \omega, \omega^2\right)$
$=-3,-3 \omega, 3 \omega^2$
$\Rightarrow x=-1,2-3 \omega, 2-3 \omega^2$
View full question & answer→MCQ 442 Marks
The value of $(8)^{1 / 3}$ is
- A
$-1+i \sqrt{3}$
- B
$-1-i \sqrt{3}$
- C
- ✓
Answer(D)
Let $(8)^{1 / 3}=x \Rightarrow x^3-8=0$
$\Rightarrow(x-2)\left(x^2+2 x+4\right)=0$
$\Rightarrow x=2,2 \omega, 2 \omega^2$
$\Rightarrow x=2,-1+ i \sqrt{3},-1- i \sqrt{3}$
View full question & answer→MCQ 452 Marks
1, $\omega$ and $\omega^2$ are the cube roots of unity, then $\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega^4\right) \ldots$ upto 8 terms is
- ✓
$2^6$
- B
$2^{10}$
- C
$2^7$
- D
$2^8$
Answer(A)
$\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega^4\right)\left(1-\omega^3+\omega^6\right)$
$\left(1-\omega^4+\omega^8\right)\left(1-\omega^5+\omega^{10}\right)\left(1-\omega^6+\omega^{12}\right)$ $\left(1-\omega^7+\omega^{14}\right)\left(1-\omega^8+\omega^{16}\right)$
$=\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega\right)(1)\left(1-\omega+\omega^2\right)$ $\left(1-\omega^2+\omega\right)(1)\left(1-\omega+\omega^2\right)\left(1-\omega^2+\omega\right)$
$=\left(1-\omega+\omega^2\right)^3\left(1-\omega^2+\omega\right)^3=(-2 \omega)^3\left(-2 \omega^2\right)^3$
$=\left(2^3 \omega^6\right)\left(2^3 \omega^3\right)=2^6$
View full question & answer→MCQ 462 Marks
If $\omega$ is a complex cube root of unity, then $(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots$ to $2_n$ factors =
Answer(B)
$(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots$.upto $2 n$ factors
$=\left(-\omega^2\right)(-\omega)(1+\omega)\left(1+\omega^2\right) \ldots$. upto 2 n factors
$=1 \cdot 1 \cdot 1 \ldots$ upto n factors $=1$
View full question & answer→MCQ 472 Marks
If $\omega(\neq 1)$ is a cube root of unity and $(1+\omega)^7=A+B \omega$, then A and B are respectively, the numbers
Answer(C)
$(1+\omega)^7=A+B \omega \Rightarrow\left(-\omega^2\right)^7=A+B \omega$
$\Rightarrow \omega^{14}=- A - B \omega$
$\Rightarrow \omega^2 \cdot \omega^{12}=-A-B \omega \Rightarrow A+B \omega+\omega^2=0$
$\Rightarrow A=1, B=1 \quad \ldots .\left[\because 1+\omega+\omega^2=0\right]$
View full question & answer→MCQ 482 Marks
If $\left(1+\omega^2\right)^m=\left(1+\omega^4\right)^m$ and $\omega$ is an imaginary cube root of unity, then least positive integral value of $m$ is
Answer(D)
We have,
$\left(1+\omega^2\right)^{ m }=\left(1+\omega^4\right)^{ m } \quad \ldots .\left[\because \omega^3=1\right]$
$\Rightarrow\left(1+\omega^2\right)^{ m }=(1+\omega)^{ m }$
$\Rightarrow(-\omega)^{ m }=\left(-\omega^2\right)^{ m }$
$\Rightarrow\left(\frac{\omega}{\omega^2}\right)^{ m }=1$
$\Rightarrow\left(\omega^2\right)^{ m }=1=\left(\omega^3\right)$
Hence, least positive integral value of $m$ is 3 .
View full question & answer→MCQ 492 Marks
If $n$ is a positive integer not a multiple of 3 , then $1+\omega^n+\omega^{2 n}=$
Answer(C)
Let $n =3 k +1$
$\omega^{ n }+\omega^{2 n }=\omega^{3 k +1}+\omega^{2(3 k +1)}$
$=\omega^{3 k } \omega+\omega^{6 k } \omega^2$
$=\left(\omega^3\right)^{ k } \cdot \omega+\left(\omega^3\right)^{2 k } \cdot \omega^2$
$=\omega+\omega^2=-1 \quad \ldots .\left[\because \omega^3=1\right]$
Hence, $1+\omega^{ n }+\omega^{2 n }=1-1=0$
View full question & answer→MCQ 502 Marks
If $\alpha$ is an imaginary cube root of unity, then for $n \in N$, the value of $\alpha^{3 n+1}+\alpha^{3 n+3}+\alpha^{3 n+5}$ is
Answer(B)
Since $\alpha$ is an imaginary cube root of unity, let it be $\omega$, then
$(\omega)^{3 n+1}+(\omega)^{3 n+3}+\omega^{3 n+5}$
$=\omega+1+\omega^5 \quad \ldots\left[\because \omega^{3 n}=1\right.$ and $\left.\omega^3=1\right]$
$=\omega+1+\omega^2=0$
View full question & answer→