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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...

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

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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

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

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  • $\begingroup$ how does this answer the question? $\endgroup$ – Dhruv Garg Dec 7 '19 at 10:26
  • $\begingroup$ @DhruvGarg Let consider carefully the meaning of the last expression. What is its real part? $\endgroup$ – user Dec 7 '19 at 10:27
  • $\begingroup$ I don't get it. I'm not aware of the symbol that you've use in the numerator ('t' like symbol) $\endgroup$ – Dhruv Garg Dec 7 '19 at 10:35
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    $\begingroup$ @DhruvGarg Ah ok sorry, it indicates the imaginary part, indeed recall that $$z-\bar z=2i \Im(z)$$ that is by $$z=x+iy \implies z-\bar z=2y$$ $\endgroup$ – user Dec 7 '19 at 10:36
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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 $0$

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