If $\left | \frac{z_1 + z_2}{z_1 - z_2}\right | = 1$, then what is $z_1\over z_2$

Question: $z_1,z_2$ are two complex numbers with $z_2 \neq 0$ and $z_1 \neq z_2$ and satisfying: $$\left | \frac{z_1 + z_2}{z_1 - z_2}\right | = 1$$ Then $z_1\over z_2$ is:

A) Real and negative

B) Real and Positive

C) Purely imaginary

D) None of the above

I think the answer should be (D) as we are only given the condition that $z_2 \neq 0$. If both $z_1,z_2 \neq 0$ then the answer would have been (C). Am I right?

• My main concern here is whether $0$ can be considered to be purely imaginary or not. – Gummy bears May 7 '16 at 13:58

If $a=\dfrac{z_1}{z_2}=x+iy$
we have $|a-1|=|a+1|\iff (x-1)^2=(x+1)^2\iff x=0$
• I understand this. However, what if $z_1 = 0$? In that case we get the value of $a$ as zero. Is that considered to be purely imaginary? – Gummy bears May 7 '16 at 13:43
A more lenghty approach, with $z_1=a_1+b_1\cdot i$ and $z_2=a_2+b_2\cdot i$:
\begin{align} \frac{|z_1+z_2|}{|z_1-z_2|} & = 1 \newline |z_1+z_2| & = |z_1-z_2| \newline \sqrt{(a_1+a_2)^2+(b_1+b_2)^2} & = \sqrt{(a_1-a_2)^2+(b_1-b_2)^2} \newline (a_1+a_2)^2+(b_1+b_2)^2 & = (a_1-a_2)^2+(b_1-b_2)^2 \newline a_1^2+2a_1a_2+a_2^2+b_1^2+2b_1b_2+b_2^2 & = a_1^2-2a_1a_2+a_2^2+b_1^2-2b_1b_2+b_2^2 \newline 2a_1a_2+2b_1b_2 & = -2a_1a_2-2b_1b_2 \newline 4a_1a_2+4b_1b_2 & = 0 \newline a_1a_2+b_1b_2 & = 0 \newline \end{align} So, let's look at $\frac{z_1}{z_2}$: $$\frac{z_1}{z_2} = \frac{a_1+b_1\cdot i}{a_2+b_2\cdot i} = \left(\frac{a_1a_2+b_1b_2}{a_2^2+b_2^2}\right)+\left(\frac{b_1a_2-a_1b_2}{a_2^2+b_2^2}\right)\cdot i \\$$ Plugging in $a_1a_2+b_1b_2 = 0$ leads to: $$\left(\frac{0}{a_2^2+b_2^2}\right)+\left(\frac{b_1a_2-a_1b_2}{a_2^2+b_2^2}\right)\cdot i = 0+\left(\frac{b_1a_2-a_1b_2}{a_2^2+b_2^2}\right)\cdot i$$ So, answer C) is correct.