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

A
$(2 - 1 / \sqrt{2}) (1 - i)$
B
$(2 - 1 / \sqrt{2}) (1 + i)$
C
$(2 + 1 / \sqrt{2}) (1 - i)$
D
$(2 + 1 / \sqrt{2}) (1 + i)$

Solution:

We have,

$\begin{array}{c} | z - 2 + 2 i | = 1 \\ \Rightarrow | z - (2 - 2 i) | = 1 \end{array}$

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 $π / 4$ with OX and

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

So, coordinates of P are $[(2 \sqrt{2} - 1) \cos (π / 4), - (2 \sqrt{2} - 1) \sin (π / 4)]$ , i.e., $((2 - 1 / \sqrt{2}), - (2 - 1 / \sqrt{2}))$ .

Hence,

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

Solution:

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