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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.6$
Question 1.
If $\mathrm{z}=\mathrm{x}+$ iy is a complex number such that $\left|\frac{z-4 i}{z+4 i}\right|=1$. show that the locus of $\mathrm{z}$ is real axis.
Solution:
$
\begin{aligned}
& \left|\frac{z-4 i}{z+4 i}\right|=1 \\
& \Rightarrow|\mathrm{z}-4 \mathrm{i}|=|\mathrm{z}+4 \mathrm{i}| \\
& \text { let } \mathrm{z}=\mathrm{x}+\mathrm{iy} \\
& \Rightarrow|\mathrm{x}+\mathrm{iy}-4 \mathrm{i}|=|\mathrm{x}+\mathrm{iy}+4 \mathrm{i}| \\
& \Rightarrow|\mathrm{x}+\mathrm{i}(\mathrm{y}-4)|=|\mathrm{x}+(\mathrm{y}+4)| \\
& \Rightarrow \sqrt{x^2+(y-4)^2}=\sqrt{x^2+(y+4)^2}
\end{aligned}
$
Squaring on both sides, we get
$\mathrm{x}^2+\mathrm{y}^2-8 \mathrm{y}+16=\mathrm{x}^2+\mathrm{y}^2+16+8 y$
$\Rightarrow-16 y=0$
$\Rightarrow \mathrm{y}=0$ in two equation of real axis.
Question 2.
If $\mathrm{z}=\mathrm{x}+\mathrm{iy}$ is a complex number such that $\operatorname{Im}\left(\frac{2 z+1}{i z+1}\right)=0$ show that the locus of $\mathrm{z}$ is $2 \mathrm{x}^2+2 \mathrm{y}^2+\mathrm{x}-2 \mathrm{y}$ $=0$.
Solution:
Let $z=x+i y$
$
\begin{aligned}
\operatorname{Im}\left(\frac{2 z+1}{i z+1}\right) & =0 & \Rightarrow \operatorname{Im}\left(\frac{2(x+i y)+1}{i(x+i y)+1}\right)=0 \\
\Rightarrow \quad \operatorname{Im}\left[\frac{2 x+i 2 y+1}{i x-y+1}\right] & =0 & \Rightarrow \operatorname{Im}\left[\frac{(2 x+1)+i 2 y}{(1-y)+i x} \times \frac{(1-y)-(i x)}{(1-y)-(i x)}\right]=0
\end{aligned}
$
Considering only the imaginary parts,
$
\frac{-x(2 x+1)+2 y(1-y)}{(1-y)^2+x^2}=0 \quad \Rightarrow-2 x^2-x+2 y-2 y^2=0
$
$
2 x^2+2 y^2+x-2 y=0
$
Hence proved.

Question 3.
Obtain the Cartesian form of the locus of $z=x+$ iy in each of the following cases:
(i) $[\operatorname{Re}(\mathrm{iz})]^2=3$
(ii) $\operatorname{Im}[(1-i) z+1]=0$
(iii) $|z+i|=|z-1|$
(iv) $\bar{z}=z^{-1}$
Solution:
(i) $z=x+i y$
$[\operatorname{Re}(\mathrm{iz})]^2=3$
$\Rightarrow\left[\operatorname{Re}[\mathrm{i}(\mathrm{x}+\mathrm{iy}]]^2=3\right.$
$\Rightarrow[\operatorname{Re}(\mathrm{ix}-\mathrm{y})]^2=3$
$\Rightarrow(-\mathrm{y})^2=3$
$\Rightarrow \mathrm{y}^2=3$
(ii) $\operatorname{Im}[(1-i) z+1]=0$
$\Rightarrow \operatorname{Im}[(1-i)(z+i y)+1]=0$
$\Rightarrow \operatorname{Im}[\mathrm{x}+\mathrm{iy}-\mathrm{ix}+\mathrm{y}+1]=0$
$\Rightarrow \operatorname{Im}[(\mathrm{x}+\mathrm{y}+1)+\mathrm{i}(\mathrm{y}-\mathrm{x})]=0$
Considering only the imaginary part
$
\mathrm{y}-\mathrm{x}=0 \Rightarrow \mathrm{x}=\mathrm{y}
$
(iii) $|z+i|=|z-1|$
$\Rightarrow|\mathrm{x}+\mathrm{iy}+\mathrm{i}|=|\mathrm{x}+\mathrm{iy}-1|$
$\Rightarrow|x+i(y+1)|=|(x-1)+i y|$
Squaring on both sides
$
\begin{aligned}
& |x+i(y+1)|^2=|(x-1)+i y|^2 \\
& \Rightarrow x^2+(y+1)^2=(x-1)^2+y^2 \\
& \Rightarrow x^2+y^2+2 y+1=x^2-2 x+1+y^2 \\
& \Rightarrow 2 y+2 x=0 \\
& \Rightarrow x+y=0
\end{aligned}
$
(iv) $\bar{z}=z^{-1}$
$\Rightarrow \bar{z}=\frac{1}{z}$
$\Rightarrow z \bar{z}=1$
$\Rightarrow|z|^2=1$
$\Rightarrow|\mathrm{x}+\mathrm{iy}|^2=1$
$\Rightarrow x^2+y^2=1$

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) $|z-2-\mathrm{i}|=3$
(ii) $|2 z+2-4 i|=2$
(iii) $|3 z-6+12 i|=8$
Solution:
(i) Let $z=x+i y$ $|z-2-i|=3$
$
\begin{aligned}
& \Rightarrow|x+i y-2-i|=3 \\
& \Rightarrow|(x-2)+i(y-1)|=3 \\
& \Rightarrow \sqrt{(x-2)^2+(y-1)^2}=3
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
& (\mathrm{x}-2)^2+(\mathrm{y}-1)^2=9 \\
& \Rightarrow \mathrm{x}^2-4 \mathrm{x}+4+\mathrm{y}^2-2 \mathrm{y}+1-9=0 \\
& \Rightarrow \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-2 \mathrm{y}-4=0 \text { represents a circle } \\
& 2 \mathrm{~g}=-4 \Rightarrow \mathrm{g}=-2 \\
& 2 \mathrm{f}=-2 \Rightarrow \mathrm{f}=-1 \\
& \mathrm{c}=-4
\end{aligned}
$
(a) Centre $(-\mathrm{g},-\mathrm{f})=(2,1)=2+\mathrm{i}$
(b) Radius $=\sqrt{g^2+f^2-c}=\sqrt{4+1+4}=3$
Aliter: $|z-(2+\mathrm{i})|=3$
Centre $=2+\mathrm{i}$
radius $=3$

$
\begin{aligned}
& \text { (ii) }|2(\mathrm{x}+\mathrm{iy})+2-4 \mathrm{i}|=2 \\
& \Rightarrow|2 \mathrm{x}+\mathrm{i} 2 \mathrm{y}+2-4 \mathrm{i}|=2 \\
& \Rightarrow|(2 \mathrm{x}+2)+\mathrm{i}(2 \mathrm{y}-4)|=2 \\
& \Rightarrow|2(\mathrm{x}+1)+2 \mathrm{i}(\mathrm{y}-2)|=2 \\
& \Rightarrow|(\mathrm{x}+1)+\mathrm{i}(\mathrm{y}-2)|=1 \\
& \Rightarrow \sqrt{(x+1)^2(y-2)^2}=1
\end{aligned}
$
Squaring on both sides,
$
\begin{aligned}
& x^2+2 x+1+y^2+4-4 y-1=0 \\
& \Rightarrow x^2+y^2+2 x-4 y+4=0 \text { represents a circle } \\
& 2 g=2 \Rightarrow g=1 \\
& 2 f=-4 \Rightarrow f=-2 \\
& c=4
\end{aligned}
$
(a) Centre $(-\mathrm{g},-\mathrm{f})=(-1,2)=-1+2 \mathrm{i}$
(b) Radius $=\sqrt{g^2+f^2-c}=\sqrt{1+4-4}=1$
Aliter: $2|(z+1-2 i)|=2$
$
|z-(-1+2 \mathrm{i})|=1
$
Centre $=-1+2 \mathrm{i}$
radius $=1$

$
\begin{aligned}
& \text { (iii) }|3(\mathrm{x}+\mathrm{iy})-6+12 \mathrm{i}|=8 \\
& \Rightarrow|3 \mathrm{x}+\mathrm{i} 3 \mathrm{y}-6+12 \mathrm{i}|=8 \\
& \Rightarrow|3(\mathrm{x}-2)+\mathrm{i} 3(\mathrm{y}+4)|=8 \\
& \Rightarrow 3|(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}+4)|=8 \\
& \Rightarrow 3 \sqrt{(x-2)^2+(y+4)^2}=8
\end{aligned}
$
Squaring on both sides,
$
\begin{aligned}
& 9\left[(\mathrm{x}-2)^2+(\mathrm{y}+4)^2\right]=64 \\
& \Rightarrow \mathrm{x}^2-4 \mathrm{x}+4+\mathrm{y}^2+8 \mathrm{y}+16=\frac{64}{9} \\
& \Rightarrow \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+8 \mathrm{y}+20-\frac{64}{9}=0 \\
& \mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+8 \mathrm{y}+\frac{116}{9}=0 \text { represents a circle. } \\
& 2 \mathrm{~g}=-4 \Rightarrow \mathrm{g}=-2 \\
& 2 \mathrm{f}=8 \Rightarrow \mathrm{f}=4 \\
& \mathrm{c}=\frac{116}{9}
\end{aligned}
$
(a) Centre $(-\mathrm{g},-\mathrm{f})=(2,-4)=2-4 \mathrm{i}$
(b) Radius $==\sqrt{g^2+f^2-c}=\sqrt{4+16-\frac{116}{9}}=\sqrt{\frac{180-116}{9}}=\frac{8}{3}$
Aliter:
$
\begin{aligned}
& |z-2+4 i|=\frac{8}{3} \\
& \Rightarrow|z-(2-4 i)|=\frac{8}{3}
\end{aligned}
$
Centre $=2-4 \mathrm{i}$, Radius $=\frac{8}{3}$

Question 5.
Obtain the Cartesian equation for the locus of $z=x+$ iy in each of the following cases.
(i) $|z-4|=16$
(ii) $|z-4|^2-|z-1|^2=16$
Solution:
$
\begin{aligned}
& \text { (i) } z=x+i y \\
& |z-4|=16 \\
& \Rightarrow|x+i y-4|=16 \\
& \Rightarrow|(x-4)+i y|=16 \\
& \Rightarrow \sqrt{(x-4)^2+y^2}=16
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
& (\mathrm{x}-4)^2+\mathrm{y}^2=256 \\
& \Rightarrow \mathrm{x}^2-8 \mathrm{x}+16+\mathrm{y}^2-256=0 \\
& \Rightarrow \mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-240=0 \text { represents the equation of circle }
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) }|x+i y-4|^2-|x+i y-1|^2=16 \\
& \Rightarrow|(x-4)+i y|^2-|(x-1)+i y|^2=16 \\
& \Rightarrow\left[(x-4)^2+y^2\right]-\left[(x-1)^2+y^2\right]=16 \\
& \Rightarrow\left(x^2-8 x+16+y^2\right)-\left(x^2-2 x+1+y^2\right)=16 \\
& \Rightarrow x^2+y^2-8 x+16-x^2+2 x-1-y^2=16 \\
& \Rightarrow-6 x+15=16 \\
& \Rightarrow 6 x+1=0
\end{aligned}
$