I have come across the intriguing statement that $\frac{\bar z}{z^2}-\frac{z}{\bar z^2}$ is an imaginary number, but somehow - no matter what I try - I cannot seem to get to that result (I have tried a long list of operations, but all of the outcomes still had the 'a' of 'a+ib' or 'a-ib' in it in the end). Can anybody shed some light on this?
Thank you!
On
Note that $\frac{\overline{z}}{z^2}-\frac{z}{\overline{z}^2} = \frac{\overline{z}^3-z^3}{z^2\overline{z}^2}$ (1). You should know that $z\overline{z} = \lvert z\rvert \in \mathbb{R}$. By consequence, the denominator of (1) is real. To prove that your expression is imaginary now comes down to proving that the numerator of (1) is imaginary. From here on it should be a lot easier. The conclusion of the proof is in the spoiler.
It is now useful to write $z=a+ib$, with $a,b\in\mathbb{R}$. The numerator of (1) then becomes $(a^3-ia^2b-ab^2+ib^3)-(a^3+ia^2b-ab^2-ib^3) = -2ia^2b+2ib^3$, which is completely imaginary. We are done.
Let $$\frac{\overline{z}}{z^2} = a+ib.$$ Then $$\overline{\left(\frac{\overline{z}}{z^2}\right)} =\frac{z}{\overline{z}^2}= a-ib.$$ Consequently, $$\frac{\overline{z}}{z^2} - \frac{z}{\overline{z}^2} = ?$$