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COLUMN A
COLUMN B

a) The polar form of $i+\sqrt3$ is i) Perpendicular bisector of segment joining (-2,0) and (2,0)

b)The amplitude of $-1+\sqrt{-3}$ is ii) On or outside the circle having centre at (0,-4) and radius 3.

c) If $\left | z+2 \right |=\left | z-2 \right |$ then locus of z is iii) $2\pi/3$

d)If $\left | z+2i \right |=\left | z-2i \right |$ then locus of z is
iv)Perpendicular bisector of segment joining (0,-2) and (0,2)

e) Region represented by $\left | z+4i \right |\geq 3$ is v) $2\left ( \cos\frac{\pi}{6}+i\sin\frac{\pi}{6} \right )$

f) Region represented by $\left | z+4 \right |\geq 3$ is vi) On or outside the circle having centre at (-4,0) and radius 3units.

g)Conjugate of $\frac{1+2i}{1-i}$ lies in vii) First Quadrant

h) Reciprocal of 1-i lies in viii) Third Quadrant

Answers (1)

a)Given $z=i+\sqrt3$

Polar form of z $=r[cos\theta+isin\theta]=i+\sqrt3$ .

$r=\sqrt{3+1} =2$

And $\tan \alpha = \frac{1}{\sqrt3}$

$\alpha = \frac{\pi}{6}$

Since $x>0,y>0$

Polar form of z $=2[cos\frac{\pi}{6}+isin\frac{\pi}{6}]$

b) Given that $z=-1+\sqrt3=-1+\sqrt3 i$

Here argument z $=\tan^{-1}\left | \frac{\sqrt3}{-1} \right |= \tan^{-1}\sqrt{3}= \frac{\pi}{3}$

So, $\alpha = \frac{\pi}{3}$

Since, x<0 and y>0 $\theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

c) $|z+2|=|z-2|$

$|x+iy+2|=|x+iy-2|$

$|(x-2)+iy|=|(x-2)+iy|$

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

$(x+2)^2+y^2=(x-2)^2+y^2$

$(x+2)^2=(x-2)^2$

$x^2+4+4x=x^2+4-4x$

$8x=0 , x=0$

which represents equation of-y axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) $|z+2i|=|z-2i|$

$(x+iy+2i)=|x+iy-2i|$

$|x+(y+2)i|=|x+(y-2)i|$

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

$x^2+(y+2)^2=x^2+(y-2)^2$

$(y+2)^2=(y-2)^2$

$y^2+4+4y=y^2+4-4y$

$8y=0 ,y=0$

which is the equation of x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) $|z+4i|\geq3$

$|x+iy+4|\geq3$

$|x+(y+4)i|\geq3$

$\sqrt{x^2+(y+4)^2} \geq3$

$x^2+(y+4)^2\geq9$

$x^2+y^2+8y+16 \geq9$

$x^2+y^2+8y+7 \geq0$ $r=\sqrt{(4)^2-7}=3$

which represents a circle on or outside having centre (0,-4)

f) $|z+4|\leq3$

Let $z=x+iy$ Then, $|x+iy+4|\leq3$

$|(x+4)+iy|\leq3$

$\sqrt{(x+4)^2+y^2 } \leq3$

$x^2+8x+16+y^2\leq9$

$x^2+y^2+8x+7 \leq0$ which is a circle having centre (-4,0) and $r=\sqrt{(4)^2-7}=\sqrt9=3$ and is on the circle.

g) Let $z=\frac{1+2i}{1-i}$

$\frac{1+2i}{1-i}* \frac{1+i}{1+i}=\frac{1+i+2i+2i^2}{1-i^2}$

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

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

$=-\frac{1}{2}+\frac{3}{2}i$ which lies in third quadrant..

h) Given that z=1-i

Reciprocal of $z=\frac{1}{z}=\frac{1}{1-i}*\frac{1+i}{1+i}$

$=\frac{1+i}{1+i^2}$ which lies in first quadrant.

$So,(a)-(v),(b)-(iii),(c)-(i),(d)-(iv),(e)-(ii),(f)-(vi),(g)-(viii),(h)-(vii)$

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