prove that Z1=icotθ2Z_1 = i\cot \frac{\theta}{2} given that Z=cosθ+isinθZ = \cos \theta + i\sin \theta

If Z=cosθ+isinθZ = \cos \theta+i\sin \theta, Z1=z+1z−1Z_1= \dfrac{z+1}{z-1}, prove that Z1=−icotخ¸2Z_1= -i\cot \dfrac{خ¸}{2}\\

This was a proof that I ran into in a quiz. I couldn’t really solve it. I only got as far as Z1=z+1Z_1 = z+1, Z1=cosθ+sinθZ_1 = \cos \theta+\sin \theta −1-1.

It’s bugging me since then. A friend told me we’d prove it using a double angle formula.

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write ZZ as eiθe^{i\theta} and factor by eiθ2e^{i\frac{\theta}{2}}.
– Abdallah Hammam
2 days ago

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1

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z+1z−1=(cosθ+1)+isinθ(cosθ−1)+isinθ=2cos2θ2+2isinθ2cosθ2−2sin2θ2+2isinθ2cosθ2=cotθ2(cosθ2+isinθ2−sinθ2+icosθ2)=cotθ2cosθ2+isinθ2i(cosθ2+isinθ2)=−icotθ2\begin{align*}
\frac{z+1}{z-1} &=
\frac{(\cos \theta+1)+i\sin \theta}{(\cos \theta-1)+i\sin \theta} \\
&=
\frac{2\cos^2 \frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos \frac{\theta}{2}}
{-2\sin^2 \frac{\theta}{2}+2i\sin \frac{\theta}{2} \cos \frac{\theta}{2}}
\\ &=
\cot \frac{\theta}{2}
\left(
\frac{\cos \frac{\theta}{2}+i\sin \frac{\theta}{2}}
{-\sin \frac{\theta}{2}+i\cos \frac{\theta}{2}}
\right) \\ &=
\cot \frac{\theta}{2}
\frac{\cos \frac{\theta}{2}+i\sin \frac{\theta}{2}}
{i\left(\cos \frac{\theta}{2}+i\sin \frac{\theta}{2} \right)} \\
&= -i\cot \frac{\theta}{2}
\end{align*}

I put a negative in front of the ii in the OP
– imranfat
2 days ago