Question

If $\left(\frac{z-1}{z+1}\right)$ is purely imaginary then:

• A. $|z|=1$
• B. $|z|>1$
• C. $|z|<1$
• D. $|z|<2$

Answer 1

Answer: (A)

$|z|=1$

Let $A$ be the point denoting $z$ on the Argand plane. Let $B$ be the point $(1,0)$ and $C$ be the point $(0,-1)$

The numerator in the given expression is the vector along $BA$, the denominator is the vector along $CA$

Since, the ratio is purely imaginary, it means that $BA$ is perpendicular to $CA$.
Hence, the locus of $A$ would be a circle with centre as the mid-point of $BC$ and the diameter $=BC$. Hence, the centre is $(0,0)$ and radius $=1$.

Hence, $|z|=1$