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Let $\omega=\cos \theta + i \sin \theta$. Find, in terms of $\theta$, the argument of $(1-\omega ^2)^*$

I started by using De Moivre's theorem and making the conjugate.

Let $\alpha$ be required argument

\begin{equation*} \tan \alpha=\dfrac{\sin 2\theta}{1-\cos 2\theta}, \end{equation*}

$\tan \alpha=\cot \theta$.

I'm a bit stuck here. I looked in the markscheme and it says the argument is $\dfrac{\pi}{2}-\theta$

I kind of understand why it is, but it's not really clicking in my head yet. Can someone explain this clearly to me please?

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  • $\begingroup$ Is that little star there the conjugate of $\;1-w^2\;$ ? Because if it is I think it should be $$\tan\alpha=\frac{\sin 2\theta}{1-\cos2\theta}$$ $\endgroup$ – Timbuc May 6 '15 at 16:06
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You can either use the identity

$$\frac{\sin\alpha}{\cos\alpha}=\frac{\cos \theta}{\sin\theta}\implies \cos\theta\cos\alpha-\sin\theta\sin\alpha=0\\ \cos{(\theta+\alpha)}=0\implies \theta+\alpha=\pi/2$$

or use a triangle to see:

enter image description here

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Using the polar form:

$$w=e^{i\theta}\implies \overline{1-w^2}=\overline1-\overline{e^{2i\theta}}=1-e^{-2i\theta}=1-\cos2\theta+i\sin2\theta=$$

$$=2\cos^2\theta+2i\cos\theta\sin\theta$$

From here, and using your notation, we get

$$\tan\alpha=\frac{2\cos\theta\sin\theta}{2\cos^2\theta}=\frac{\sin\theta}{\cos\theta}=\tan(\theta+k\pi)\;,\;\;k\in\Bbb Z$$

which is the same as with $\;w\mod\pi\;$ ...and I can't see where that expression you say is from the markscheme comes from.

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