Simple complex problems !


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.


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).


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}}


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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