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i) For any two complex numbers z₁, z₂ and any real numbers a, b, $|az_1 - bz_2|^2 + |bz_1 + az_2|^2 =....$

ii) The value of $\sqrt{-25} * \sqrt{-9}$ is

iii) The number $\frac{(1-i)^3}{1-i^3}$ is equal to .......

iv) The sum of the series $i+i^2+i^3+...+i^{1000}$ upto 1000 terms is ...

v) Multiplicative inverse of 1 + i is ................

vi) If $z_1 \ and \ z_2$ are complex numbers such that $z_1 + z_2$ is a real number, then $z_2$ =

vii) $arg (z)+arg (\bar{z} )$ $(\bar{z}\neq0)$ is ....

viii) If $|z+4|\leq 3$ then the greatest and least values of |z+1| are ..... and .....

ix) if $\frac{z-2}{z+2}=\frac{\pi}{6}$ then the locus of z is .......

x) If $|z|=4$ and $arg (z) = \frac{5\pi}{6}$ then z=

Answers (1)

i) $|az_1 - bz_2|^2 + |bz_1 + az_2|^2$

$=|az_1 |^2+|bz_2 |^2-2Re(az_1.b(\bar{z_2} ) +|bz_1 |^2+|az_2 |^2+2Re(az_1.b\bar{z_2} ) ?)$

$=|az_1 |^2+|bz_2 |^2+|bz_1 |^2+|az_2 |^2=(a^2+b^2 )(|z_1 |^2+|z_2 |^2 )$

ii) $\sqrt{-25}*\sqrt{-9}=5i*3i=15i^2=-15$

iii) $\frac{(1-i)^3}{1-i^3}=\frac{(1-i)^3}{(1-i)(1+i+i^2 )}$

$=\frac{(1-i)^3}{(1+i-1) }= \frac{1+i^2-2i}{}i$

$=\frac{1-1-2i}{i}=-\frac{2i}{i}=-2$

iv) $i+i^2+i^3+...+i^{1000}=0 \left [ \sum_{n=1}^{1000}i^n=0 \right ]$

v) $\frac{1}{1+i}=1*\frac{1-i}{(1+i)(1-i)} =\frac{1}{2} (1-i)$

vi) Let $z_1=x_1+iy_1 \: \: and \: \: z_2=x_2+iy_2$

$z_1+z_2=(x_1+x_2 )+i(y_1+y_2 )$

If $z_1+z_2$ is real, then $y_1+y_2=0$

$y_1=-y_2$

$z_2=x_2-iy_1$

$z_2=x_1-iy_1 (\: \: when\: \: x_1=x_2 )$

So, $z_2=\bar{z_1}$

vii) $arg (z)+arg (\bar{z} )$

$If\: \: arg (z)= \theta , then \arg (\bar{z})=-\theta$

$\theta+(-\theta)=0$

viii) $|z+4|\leq 3$

$=|z+4-3|\leq|z+4|+|-3|$

$=|z+4-3|\leq3+3$

$=|z+4-3|\leq6$

The greatest value is 6 and the least value of $|z+1|$ is 0

ix) $\frac{z-2}{z+2}=\frac{\pi}{6}$

Let $z=x+iy$

$\left |\frac{x+iy-2}{x+iy+2} \right |=\left |\frac{(x-2)+iy}{(x+2)+iy} \right |= \frac{\pi}{6}$

$6|(x-2)+iy|=\pi|(x+2)+iy|$

$6\sqrt{(x-2)^2+y^2 }=\pi\sqrt{(x+2)^2+y^2 }$

$36[x^2+4-4x+y^2 ]=\pi ^2[x^2+4+4x+y^2 ]$

$36x^2+144-144x+36y^2=\pi^2 x^2+4\pi^2+4\pi^2 x+\pi^2 y^2$

$(36-\pi^2 ) x^2+(36-\pi^2 )-(144+4\pi^2 )x+144-4\pi^2=0$

which represents the equation of a circle

x) $|z|=4$ and $arg (z) = \frac{5\pi}{6}$

Let $z=x+iy$

$|z|= \sqrt{x^2+y^2} =4$

$x^2+y^2=16$

$arg(z)=\tan^{-1}\frac{y}{x}=\frac{5\pi}{6}$

$\frac{y}{x}=\tan\left ( \pi-\frac{\pi}{6} \right )$

$=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$

$x=-\sqrt3 y$

$(-\sqrt3 y)^2+y^2=16$

$3y^2+y^2=16$

$4y^2=16$

$y^2=4$

$y=\pm2$

$x=-2\sqrt3$

$z=-2\sqrt3+2i$

Posted by

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