Complex numbers, set of values for which z will be purely real or imaginary.

Question

A complex number z is given by $ z = \frac{a+i}{a-i}, a∈R$.

• Determine the set of values of a such that

(a). z is purely real;

(b). z is purely imaginary.

(c). Show that |z| is a constant for all values of a.

Hi all,

I solved the question partially - I managed to find that the number will be purely imaginary for a=1, and a=- 1. However, when I try to solve for a purely real value, a is undefined. I think I solved part b, but the proof wasn't all that clear cut and I was hoping you could clarify that a bit for me. Thanks,

Answer 1

Consider $w=a+i$, then $z=\frac{w}{\bar{w}}=\frac{w^2}{w\bar{w}}=\frac{w^2}{|w|^2}$. Since $|w|^2=1+a^2 \in \Bbb{R}$ so you only have to deal with the numerator.

Now $w^2=(a^2-1)+2ai$.

1. For $z$ to be purely imaginary, we want $a^2-1=0 \implies a=\pm 1$.
2. For $z$ to be purely real, we want $2a=0 \implies a=0$.
3. For $|z|$ we have $|z|=\left|\frac{w^2}{|w|^2}\right|=\frac{|w^2|}{|w|^2}=\frac{|w|^2}{|w|^2}=1.$ for all $a$.

Answer 2

hint

$$z=\frac{a+i}{a-i}=\frac{(a+i)(a+i)}{(a-i)(a+i)}$$

$$=\frac{a^2-1}{a^2+1}+\frac{2ia}{a^2+1}$$

$z$ is purely real if $2a=0$.

$$|z|^2=\frac{(a+i)(a-i)}{(a-i)(a+i)}=1$$