When complex numbers get angry....

Number Theory Level 2

If z z and w w are complex numbers such that ∣ z ∣ = ∣ w ∣ = 1 |z|=|w|=1 and z w ≠ − 1 zw \ne -1 ,

then what kind of number will z + w z w + 1 \large \dfrac{z + w}{zw +1} be?

Neither real nor purely imaginary Real number Multiple of 1 + i 1 + i Purely imaginary

Syed Hamza Khalid by Syed Hamza Khalid , 17, France

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1 solution

Chris Lewis
Oct 30, 2019

Let z = e i a z=e^{ia} and w = e i b w=e^{ib} where a , b a,b are real numbers.

The ratio is then z + w z w + 1 = e i a + e i b e i ( a + b ) + 1 = ( e − i ( a + b ) + 1 ) ( e i a + e i b ) ( e − i ( a + b ) + 1 ) ( e i ( a + b ) + 1 ) = e i a + e − i a + e i b + e − i b 2 + e i ( a + b ) + e − i ( a + b ) = cos ⁡ a + cos ⁡ b 1 + cos ⁡ ( a + b ) \begin{aligned} \frac{z+w}{zw+1} &= \frac{e^{ia}+e^{ib}}{e^{i(a+b)}+1} \\ &=\frac{\left(e^{-i(a+b)}+1\right) \left(e^{ia}+e^{ib} \right)}{\left(e^{-i(a+b)}+1\right)\left(e^{i(a+b)}+1\right)} \\ &=\frac{e^{ia}+e^{-ia}+e^{ib}+e^{-ib}}{2+e^{i(a+b)}+e^{-i(a+b)}} \\ &=\frac{\cos{a}+\cos{b}}{1+\cos{(a+b)}} \end{aligned}

which is real, as long as a + b ≠ ( 2 k + 1 ) π a+b \neq (2k+1)\pi for any integer k k ; this condition is equivalent to the given condition z w ≠ − 1 zw \neq -1 , so we're done.

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