## Simple complex problems !

**Question:**

If z lies on the circle , then equals

(a) 0

(b) 2

(c) -1

(d) none of these.

**Solution:**

Note that represents a circle with the segment joining and as a diameter. (*draw this circle for yourself!)*

If z lies on this circle, then is purely imaginary.

Ans. (d).

**Question:**

If and (where z \neq -1), then equals

(a) 0

(b)

(c)

(d)

**Solution:**

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, .

More later,

Nalin Pithwa

### Like this:

Like Loading...

*Related*