Exercise 2.6-Additional Problems - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
If the imaginary part of $\frac{2 z+1}{i z+1}$ is $-2$, then show that the locus of the point representing $\mathrm{z}$ in the argand plane is a straight line.
Solution:
Let $z=x+i y$. Then,
$
\begin{aligned}
& \begin{aligned}
& \frac{2 z+1}{i z+1}=\frac{2(x+i y)+1}{i(x+i y)+1}=\frac{(2 x+1)+i 2 y}{(1-y)+i x} \\
&=\frac{(2 x+1)+i 2 y}{(1-y)+i x} \times \frac{(1-y)-i x}{(1-y)-i x}=\frac{(2 x+1-y)+i\left(2 y-2 y^2-2 x^2-x\right)}{(1-y)^2+x^2} \\
&=\left\{\frac{2 x+1-y}{x^2+(1-y)^2}\right\}+i\left\{\frac{2 y-2 y^2-2 x^2-x}{x^2+(1-y)^2}\right\} \\
& \therefore \operatorname{Im}\left(\frac{2 z+1}{i z+1}\right)=\left(\frac{2 y-2 y^2-2 x^2-x}{x^2+(1-y)^2}\right\} \\
& \text { But, it is given that } \operatorname{Im}(z)=-2
\end{aligned} \\
& \Rightarrow \quad 2 y-2 y^2-2 x^2-x=-2 x^2-2(1-y)^2 \\
& \Rightarrow \quad x+2 y-2=0, \text { which is a straight line. }
\end{aligned}
$
Hence, the locus of $\mathrm{z}$ is a straight line

Question 2.

If the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is 4 , then show that locus of the point representing $\mathrm{z}$ in the complex plane is a circle.
Solution:
Let $z=x+i y$. Then, $\bar{z}=x-i y$
$
\begin{aligned}
\therefore \quad \frac{\bar{z}+2}{\bar{z}-1} & =\frac{x-i y+2}{x-i y-1}=\frac{(x+2)-i y}{(x-1)-i y}=\frac{(x+2)-i y}{(x-1)-i y} \times \frac{(x-1)+i y}{(x-1)+i y} \\
& =\frac{\left(x^2+y^2+x-2\right)+3 i y}{(x-1)^2+y^2}=\left\{\frac{x^2+y^2-x-2}{(x-1)^2+y^2}\right\}+i\left\{\frac{3 y}{(x-1)^2+y^2}\right\}
\end{aligned}
$
It is given that the real part of $\frac{\bar{z}+2}{\bar{z}-1}$ is 4 .
$
\begin{aligned}
& \therefore \quad \frac{x^2+y^2-x-y}{(x-1)^2+y^2}=4 \\
& \Rightarrow 3 x^2+3 y^2-7 x+y+4=0 \text {, which represents a circle. } \\
&
\end{aligned}
$

Question 3.
Show that $\left|\frac{z-2}{z-3}\right|=2$ represents a circle. Find its centre and radius.
Solution:
Let
Then,
$
\begin{aligned}
& \Rightarrow \quad|(x-2)+i y|=2|(x-3)+i y| \\
& \Rightarrow \quad \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\
& \Rightarrow \quad(x-2)^2+y^2=4\left[(x-3)^2+y^2\right] \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\
& \Rightarrow x^2+y^2-\frac{20}{3} x+\frac{32}{3}=0 \\
& \Rightarrow\left(x-\frac{10}{3}\right)^2+(y-0)^2=\left(\frac{2}{3}\right)^2 \text {, } \\
&
\end{aligned}
$
which represents a circle with centre at $(10 / 3,0)$ and radius $2 / 3$
Question 4.
If $\arg (z-1)=\frac{\pi}{6}$ and $\arg (z+1)=2 \frac{\pi}{3}$, then prove that $|z|=1$.
Solution:

Given $\arg (z-1)=\frac{\pi}{6}=30^{\circ}$ and $\arg (z+1)=\frac{2 \pi}{3}=120^{\circ}$
So, $\arg (z+1)-\arg (z-1)=120^{\circ}-30^{\circ}=90^{\circ}$
$
\begin{aligned}
& \text { i.e., } \arg \frac{z+1}{z-1}=90^{\circ}=\frac{\pi}{2} \\
& \Rightarrow \operatorname{Re} \frac{z+1}{z-1}=0 \quad\left[\therefore \arg =\tan \theta=\frac{\mathrm{IP}}{\mathrm{RP}}\right] \\
& \text { Let } z=x+i y \\
& \text { So, } \frac{z+1}{z-1}=\frac{x+i y+}{x+i y-}=\frac{(x+1)+i y}{(x-1)+i y}=\frac{(x+1)+i y}{(x-1)+i y} \times \frac{(x-1)-i y}{(x-1)-i y} \\
& =\frac{(x+1)(x-1)+y^2-i y(x+1)+i y(x-1)}{(x-1)^2-y^2} \\
& \therefore \quad \operatorname{RP} \text { of } \frac{z+1}{z-1}=0 ; \frac{x^2-1+y^2}{(x-1)^2+y^2}=0 \\
& \Rightarrow \quad x^2-1+y^2=0 \quad \Rightarrow x^2+y^2=1 \\
& \text { i.e., } \quad|z|^2=1 \quad \Rightarrow|z|=1 \\
&
\end{aligned}
$
Question 5.
$P$ represents the variable complex number $z$. Find the locus of $P$, if $
\operatorname{Im}\left(\frac{2 z+1}{i z+1}\right)=-2
$

Solution:
Let
$
\begin{aligned}
& \therefore \quad \frac{2 z+1}{i z+1}=\frac{2(x+i y)+1}{i(x+i y)+1}=\frac{(2 x+1)+i(2 y)}{i x-y+1} \\
& =\frac{(2 x+1)+i(2 y)}{(1-y)+i x} \times \frac{(1-y)-i x}{(1-y)-i x} \\
& =\frac{(2 x+1)(1-y)+i 2 y(1-y)-i x(2 x+1)+x(2 y)}{(1-y)^2+x^2} \\
& \text { IP of } \frac{2 z+1}{i z+1}=-2 \\
& \Rightarrow \frac{2 y(1-y)-x(2 x+1)}{(1-y)^2-x^2}=-2 \\
& \Rightarrow \quad 2 y-2 y^2-2 x^2-x=-2\left[1+y^2-2 y+x^2\right] \\
& \text { i.e., } \quad-2 y^2-2 x^2-x+2 y+2+2 y^2-4 y+2 x^2=0 \\
& -x-2 y+2=0 \\
& \Rightarrow \quad x+2 y-2=0 \\
& \text { i.e., } x+2 y=2 \\
&
\end{aligned}
$

Question 6.
$P$ represents the variable complex number $z$. Find the locus of $P$, if
$
\operatorname{Re}\left(\frac{z+1}{z+i}\right)=1
$
Solution:
Let $z=x+i y$
So,
$
\begin{aligned}
& \frac{z+1}{z+i}=\frac{x+i y+1}{x+i y+i}=\frac{(x+1)+i y}{x+i(y+1)}=\frac{(x+1)+i y}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)} \\
& =\frac{x(x+1)+y(y+1)+i x y-i(x+1)(y+1)}{x^2+(y+1)^2} \\
& \text { RP of }\left(\frac{z+1}{z+i}\right)=1 \quad \Rightarrow \frac{x(x+1)+y(y+1)}{x^2+(y+1)^2}=1 \\
&
\end{aligned}
$
$
\begin{aligned}
& \text { i.e., } x(x+1)+y(y+1)=x^2+(y+1)^2 \\
& x^2+x+y^2+y=x^2+y^2+2 y+1 \\
& \Rightarrow x+y-2 y-1=0 ; \quad \text { i.e., } x-y-1=0 \\
&
\end{aligned}
$