Question 14 Marks
The conjugate of a complex number $z$, is the complex number, obtained by changing the
sign of imaginary part of $z$ . It is denoted by $\bar{z}$.
The modulus $($or absolute value$)$ of a complex number, $z = a +$ ib is defined as the non
negative real number
$\sqrt{a^2+b^2}$. It is denoted by $|z|$. i.e.
$|z|=\sqrt{a^2+b^2}$
Multiplicative inverse of $z$ is $\frac{\bar{z}}{|z|^2}$. It is also called reciprocal of $z .$
$z \bar{z}=|z|^2$
$i.$ If $f(z)=\frac{7-z}{1-z^2}$, where $z=1+2 i$, then find $|f(z)|. ( 1 )$
$ii.$ Find the value of $(z+3)(\bar{z}+3). (1)$
$iii.$ If ( $x-i y)(3+5 i)$ is the conjugate of $-6-24 i$, then find the value of $x+y. (2)$
OR
If $z=3+4 i$, then find $\bar{z}. (2)$
sign of imaginary part of $z$ . It is denoted by $\bar{z}$.
The modulus $($or absolute value$)$ of a complex number, $z = a +$ ib is defined as the non
negative real number
$\sqrt{a^2+b^2}$. It is denoted by $|z|$. i.e.
$|z|=\sqrt{a^2+b^2}$
Multiplicative inverse of $z$ is $\frac{\bar{z}}{|z|^2}$. It is also called reciprocal of $z .$
$z \bar{z}=|z|^2$
$i.$ If $f(z)=\frac{7-z}{1-z^2}$, where $z=1+2 i$, then find $|f(z)|. ( 1 )$
$ii.$ Find the value of $(z+3)(\bar{z}+3). (1)$
$iii.$ If ( $x-i y)(3+5 i)$ is the conjugate of $-6-24 i$, then find the value of $x+y. (2)$
OR
If $z=3+4 i$, then find $\bar{z}. (2)$
Answer
View full question & answer→$i.$ Let $z = 1 + 2i$
$\Rightarrow|z|=\sqrt{1+4}=\sqrt{5}$
Now $, f ( z )=\frac{7-z}{1-z^2}=\frac{7-1-2 i}{1-(1+2 i)^2}$
$=\frac{6-2 i}{1-1-4 i 2-4 i}=\frac{6-2 i}{4-4 i}$
$=\frac{(3-i)(2+2 i)}{(2-2 i ) (2+2 i)}$
$=\frac{6-2 i+6 i-2 i^2}{4-4 i^2}=\frac{6+4 i+2}{4+4}$
$=\frac{8+4 i}{8}=1+\frac{1}{2} i$
$f(z)=1+\frac{1}{2}$
$\therefore|f( z )|=\sqrt{1+\frac{1}{4}}$
$=\sqrt{\frac{4+1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$
Given that : $( z +3)(\bar{z}+3)$
Let $z = x + yi$
So $( z +3)(\bar{z}+3)=( x + yi +3)( x - yi +3)$
$=[(x+3)+y i][(x+3)-y i]$
$=(x+3)^2-y^2 i^2$
$=(x+3)^2+y^2$
$=|x+3+i y|^2$
$=|z+3|^2$
$iii.$ The conjugate of $-6 - 24i$ is $-6 + 24i.$
It is given that $-6 + 24i = (x – iy) (3 + 5i)$
$-6+24 i=3 x+5 x i-3 i y-5 y i^2$
$-6 + 24i = (3x + 5y) + i(5x - 3y)$
Comparing the real and imaginary parts,
$3x + 5y = -6 $
$5x - 3y = 24$
Solving these two equations we get $x = 3$ and $y = -3.$
Therefore, $x = 3$ and $y = -3$
Then $x + y = 3 - 3 = 0$
OR
$z = 3 + 4i$
$\Rightarrow \bar{z}=3-4 i$
$\Rightarrow|z|=\sqrt{1+4}=\sqrt{5}$
Now $, f ( z )=\frac{7-z}{1-z^2}=\frac{7-1-2 i}{1-(1+2 i)^2}$
$=\frac{6-2 i}{1-1-4 i 2-4 i}=\frac{6-2 i}{4-4 i}$
$=\frac{(3-i)(2+2 i)}{(2-2 i ) (2+2 i)}$
$=\frac{6-2 i+6 i-2 i^2}{4-4 i^2}=\frac{6+4 i+2}{4+4}$
$=\frac{8+4 i}{8}=1+\frac{1}{2} i$
$f(z)=1+\frac{1}{2}$
$\therefore|f( z )|=\sqrt{1+\frac{1}{4}}$
$=\sqrt{\frac{4+1}{4}}=\frac{\sqrt{5}}{2}=\frac{|z|}{2}$
Given that : $( z +3)(\bar{z}+3)$
Let $z = x + yi$
So $( z +3)(\bar{z}+3)=( x + yi +3)( x - yi +3)$
$=[(x+3)+y i][(x+3)-y i]$
$=(x+3)^2-y^2 i^2$
$=(x+3)^2+y^2$
$=|x+3+i y|^2$
$=|z+3|^2$
$iii.$ The conjugate of $-6 - 24i$ is $-6 + 24i.$
It is given that $-6 + 24i = (x – iy) (3 + 5i)$
$-6+24 i=3 x+5 x i-3 i y-5 y i^2$
$-6 + 24i = (3x + 5y) + i(5x - 3y)$
Comparing the real and imaginary parts,
$3x + 5y = -6 $
$5x - 3y = 24$
Solving these two equations we get $x = 3$ and $y = -3.$
Therefore, $x = 3$ and $y = -3$
Then $x + y = 3 - 3 = 0$
OR
$z = 3 + 4i$
$\Rightarrow \bar{z}=3-4 i$