## Simple complex problems !

Question:

If z lies on the circle $|z-1|=1$, then $\frac{z-2}{z}$ equals

(a) 0

(b) 2

(c) -1

(d) none of these.

Solution:

Note that $|z-1|=1$ represents a circle with the segment joining $z=0$ and $z=2+0i$ as a diameter. (draw this circle for yourself!)

If z lies on this circle, then $\frac{z-2}{z-0}$ is purely imaginary.

Ans. (d).

Question:

If $|z|=1$ and $w=\frac{z-1}{z+1}$ (where z \neq -1), then $\Re (w)$ equals

(a) 0

(b) $\frac{-1}{|z+1|^{2}}$

(c) $|\frac{z}{z+1}|\frac{1}{|z+1|^{2}}$

(d) $\frac{\sqrt{2}}{|z+1|^{2}}$

Solution: $w=\frac{z-1}{z+1} \Longrightarrow wz+w=z-1 \Longrightarrow w+1=z(1-w) \Longrightarrow z=\frac{1+w}{1-w} \Longrightarrow |\frac{1+w}{1-w}|=|z|=1 \Longrightarrow |1+w|=|1-w|$

This shows that w is equidistant from -1 and 1. Hence, w lies on the perpendicular bisector of the segment joining -1 and 1, that is, w lies on the imaginary axis.

Hence, $\Re (w)=0$.

More later,

Nalin Pithwa

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