# Find the value of the real part of (z+w)/(z-w) when |z| = |w|

It's an IBDP question. When I try to solve it, I end up with a fraction with many variables.

I was thinking maybe I should rationalize the fraction...

• What does IDBP mean? – kimchi lover Dec 8 '19 at 0:27
• IBDP is the acronym for the International Baccalaureate Diploma Program – Dhruv Garg Dec 15 '19 at 7:02

Write $$z=re^{it}$$ and $$w=re^{iu}$$. Then $$\frac{z+w}{z-w}=\frac{e^{it}+e^{iu}}{e^{it}-e^{iu}} =\frac{e^{i(t-u)/2}+e^{-i(t-u)/2}}{e^{i(t-u)/2}-e^{-i(t-u)/2}} =\frac{2\cos\frac12(t-u)}{2i\sin\frac12(t-u)}=i\cot\theta$$ where $$\theta=\frac12(u-t)$$ is half the angle between $$z$$ and $$w$$ in the Argand diagram. So zero real part, but the imaginary part is interesting.

We have that

$$\frac{z+w}{z-w}=\frac{z+w}{z-w}\frac{\bar z-\bar w}{\bar z-\bar w}=\frac{\bar zw-z\bar w}{|z-w|^2}$$

and

$$\bar zw-z\bar w=\bar zw-\overline{ \bar zw}=2i \Im(\bar zw)$$

therefore

$$\frac{z+w}{z-w} =2i \frac{\Im(\bar zw)}{|z-w|^2}$$

• how does this answer the question? – Dhruv Garg Dec 7 '19 at 10:26
• @DhruvGarg Let consider carefully the meaning of the last expression. What is its real part? – user Dec 7 '19 at 10:27
• I don't get it. I'm not aware of the symbol that you've use in the numerator ('t' like symbol) – Dhruv Garg Dec 7 '19 at 10:35
• @DhruvGarg Ah ok sorry, it indicates the imaginary part, indeed recall that $$z-\bar z=2i \Im(z)$$ that is by $$z=x+iy \implies z-\bar z=2y$$ – user Dec 7 '19 at 10:36

Geometrically, if we assume $$z$$ and $$w$$ are the edges of a quadrilateral with $$O$$ as a common vertex, then $$z+w$$ and $$z-w$$ are the two diagonals of that quadrilateral. Because $$|z|=|w|$$ the quadrilateral is a losenge. The diagonals of a losenge are orthogonal and therefore $${z+w\over z-w}$$ is a pure imaginary number. Its real part is $$0$$