If z is a complex number having the least abs | vedclass

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Question

(A) We have,$|z - 2 + 2i| = 1$,which can be written as $|z - (2 - 2i)| = 1$.This represents a circle in the complex plane with center $C$ at $(2, -2)$

Vedclass · Accepted Answer

$\left( 2 - \frac{1}{\sqrt{2}} \right)(1 - i)$

Vedclass · Answer

(A) We have,$|z - 2 + 2i| = 1$,which can be written as $|z - (2 - 2i)| = 1$.This represents a circle in the complex plane with center $C$ at $(2, -2)$ and radius $r = 1$.The complex number $z$ with the least absolute value corresponds to the point $P$ on the circle that is closest to the origin $O(0, 0)$.The point $P$ lies on the line segment $OC$,where $O$ is the origin and $C$ is the center $(2, -2)$.The distance $OC = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.The distance $OP = OC - CP = 2\sqrt{2} - 1$.The angle $	heta$ that $OC$ makes with the positive $x$-axis is $-\pi/4$ (since it lies in the fourth quadrant).The coordinates of $P$ are $(OP \cos 	heta, OP \sin 	heta) = ((2\sqrt{2} - 1) \cos(-\pi/4), (2\sqrt{2} - 1) \sin(-\pi/4))$.$P = ((2\sqrt{2} - 1) \frac{1}{\sqrt{2}}, (2\sqrt{2} - 1) (-\frac{1}{\sqrt{2}})) = (2 - \frac{1}{\sqrt{2}}, -(2 - \frac{1}{\sqrt{2}}))$.Thus,$z = (2 - \frac{1}{\sqrt{2}}) - i(2 - \frac{1}{\sqrt{2}}) = (2 - \frac{1}{\sqrt{2}})(1 - i)$.

If $z$ is a complex number having the least absolute value and $|z - 2 + 2i| = 1$,then $z =$

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