If z is a complex number having least modulus and |z – 2 + 2i| = 1, then z =

# If z is a complex number having least modulus and |z - 2 + 2i| = 1, then z =

1. A

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

2. B

$\left(2-1/\sqrt{2}\right)\left(1+\mathrm{i}\right)$

3. C

$\left(2+1/\sqrt{2}\right)\left(1-\mathrm{i}\right)$

4. D

$\left(2+1/\sqrt{2}\right)\left(1+\mathrm{i}\right)$

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### Solution:

We have,

$\begin{array}{c}|\mathrm{z}-2+2\mathrm{i}|=1\\ ⇒|\mathrm{z}-\left(2-2\mathrm{i}\right)|=1\end{array}$

Hence, z lies on a circle having center at (2, -2) and radius 1. It is evident from the figure that the required complex number z is given by the point P.

We find that OP makes an angle $\mathrm{\pi }/4$ with OX and

$\begin{array}{c}\mathrm{OP}=\mathrm{OC}-\mathrm{CP}\\ =\sqrt{{2}^{2}+{2}^{2}}-1=2\sqrt{2}-1\end{array}$

So, coordinates of P are , i.e., $\left(\left(2-1/\sqrt{2}\right),-\left(2-1/\sqrt{2}\right)\right)$.

Hence,

$\mathrm{z}=\left(2-\frac{1}{\sqrt{2}}\right)+\left\{-\left(2-\frac{1}{\sqrt{2}}\right)\right\}\mathrm{i}=\left(2-\frac{1}{\sqrt{2}}\right)\left(1-\mathrm{i}\right)$

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