Question

Match the statements in column-I with those I column-II [Note : Here z takes the values in the complex plane and Im(z) and Re(z) denote, respectively, the imaginary part and the real part of z]

Answer 1

A) Let $z=x+iy$
$|z-i|z||=|z+i|z||$

$|x+iy-i\sqrt{x^2+y^2}|=|x+iy+i\sqrt{x^2+y^2}|$

$|x+i(y-\sqrt{x^2+y^2})|=|x+i(y+\sqrt{x^2+y^2})|$

$\sqrt{x^2+(y-\sqrt{x^2+y^2}{)}^2}=\sqrt{x^2+(y+\sqrt{x^2+y^2}{)}^2}$
Squaring both sides, we get

$x^2+(y-\sqrt{x^2+y^2}{)}^2=x^2+(y+\sqrt{x^2+y^2}{)}^2$

$\therefore 4y\sqrt{x^2+y^2}=0\Rightarrow y=0$
Imaginary part is $0$

B) $|z+4|+|z-4|=10$
It is an ellipse with $2a=10$
$a=5$
$ae=4$
$e=\frac{4}{5}$
C) $\omega =2(cos\theta +isin\theta )$
$z=2(cos\theta +isin\theta )-1/\left(2\right(cos\theta +isin\theta \left)\right)$
$z=2(cos\theta +isin\theta )-(cos\theta -isin\theta )/2$

$z=\frac{3cos\theta +5isin\theta }{2}$
Let $z=x+iy$
So, $x=\frac{3cos\theta }{2},y=\frac{5sin\theta }{2}$
$2x/3=cos\theta ,2y/5=sin\theta$
$\Rightarrow \frac{x^2}{9/4}+\frac{y^2}{25/4}=1$
$e=\sqrt{1-\frac{b^2}{a^2}}$
$e=\frac{4}{5}$
$|z|=\sqrt{(3cos\theta /2{)}^2+(5sin\theta /2{)}^2}=\sqrt{\frac{9co{s}^2\theta }{2}+\frac{25si{n}^2\theta }{4}}=\sqrt{\frac{9+16si{n}^2\theta }{4}}$
Maximum value of $si{n}^2\theta =1$
$\Rightarrow \sqrt{\frac{9+16si{n}^2\theta }{4}}\le \frac{5}{2}$
$|Rez|\le \frac{3}{2}$
D) $z=cos\theta +isin\theta$
$z=cos\theta +isin\theta +1/(cos\theta +isin\theta )$
$z=cos\theta +isin\theta +cos\theta -isin\theta$
$z=2cos\theta$
Hence, Imaginary part is $0$
$|z|\le 2$
$Imz\le 1$