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I have an answer for this question, but I'm not very confident with it:

$\left | 1-e^{i\theta } \right |^{2}$ i.e. find the square of the modulus of 1-$e^{i\theta }$

$e^{i\theta } = R(cos\theta +isin\theta )$, R=1

= $cos\theta +isin\theta$

In cartesian form:

$a = Rcos\theta \Rightarrow cos\theta$

$b = Rsin\theta \Rightarrow sin\theta$

$\therefore 1-e^{i\Theta } = 1-cos\theta-isin\theta$

$\Rightarrow \left | 1-e^i\theta \right | = \left | 1-cos\theta-isin\theta \right |$

Where the real part, a, = $1-cos\theta$ and the imaginary, b, = $-sin\theta$

The modulus of a complex number being $\sqrt{a^{2} +b^{2}}$

$\therefore \left | 1-cos\theta-isin\theta \right | = \sqrt{(1-cos\theta)^{2} + (-sin\theta)^{2}}$

$\Rightarrow \left | 1-cos\theta-isin\theta \right |^{2} = (1-cos\theta)^{2} + (-sin\theta)^{2} \Rightarrow 1-2cos\theta +cos^{2}\theta +sin^{2}\theta \Rightarrow (\because cos^{2}\theta +sin^{2}\theta = 1) = 2-2cos\theta$

Is this correct? Or is my approach incorrect? Thanks in advance for any feedback/help :)

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  • $\begingroup$ It is correct but extremely long. I think anyone dealing with this could be dispensed with all those explanations, and then you could simply write:$$|1-\cos\theta-i\sin\theta|^2=(1-\cos\theta)^2+\sin^2\theta=2(1-\cos\theta)$$ $\endgroup$ – DonAntonio May 13 '17 at 9:20
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Your argument is lengthy, but sound.

If just the square of the modulus is needed, you can consider that $|z|^2=z\bar{z}$, so $$ |1-e^{i\theta}|^2=(1-e^{i\theta})(1-e^{-i\theta})= 1-(e^{i\theta}+e^{-i\theta})+1=2-2\cos\theta $$

If also the argument is needed, the trick with $1-e^{i\theta}$ (but also $1+e^{i\theta}$) is to set $$ \theta=2\alpha $$ and rewrite $$ 1-e^{i\theta}=1-e^{2i\alpha}= -e^{i\alpha}(e^{i\alpha}-e^{-i\alpha})=-2i(\sin\alpha)e^{i\alpha} $$ Since $0\le\theta<2\pi$, we have $0\le\alpha<\pi$, so $\sin\alpha\ge0$. Hence the modulus is $2\sin\alpha$ and, from $-i=e^{3i\pi/2}$, the argument is $$ \alpha+\frac{3\pi}{2} $$ (up to an integer multiple of $2\pi$).

Thus the square of the modulus is $$ 4\sin^2\alpha=4\sin^2\frac{\theta}{2}=4\frac{1-\cos\theta}{2}= 2(1-\cos\theta) $$

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  • $\begingroup$ Thanks for your response! That makes a lot of sense too. I love how there's always several ways to solve a question! $\endgroup$ – Skylineblue May 14 '17 at 1:46
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HINT:

Now $2-2\cos\theta=4\sin^2\dfrac\theta2$ using $\cos2A=1-2\sin^2A$

Finally for real $u,\sqrt{u^2}=|u|=\begin{cases} +u &\mbox{if } u\ge0 \\ -u & \mbox{if } u<0 \end{cases} $

Alternatively,

$$1-e^{i\theta}=-e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)=-2ie^{i\theta/2}\sin\dfrac\theta2$$

Alternatively, $$1-\cos\theta-i\sin\theta=2\sin^2\dfrac\theta2--2i\cos\dfrac\theta2\sin\dfrac\theta2=-2i\sin\dfrac\theta2\left(\cos\dfrac\theta2+i\sin\dfrac\theta2\right)$$

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  • $\begingroup$ I actually got to the 4sin^2(theta/2), but I didn't add it to the post as I figured the two were equivalent so if one was correct the other is as well. I don't quite understand why you wrote the lines afterwards about u, (u^2)^1/2 = |u| etc. Is this necessary? i.e. is simply writing 4sin^2(theta/2) the correct answer? $\endgroup$ – Skylineblue May 13 '17 at 9:20
  • $\begingroup$ @Skylineblue, $$4\sin^2\dfrac\theta2$$ is actually the square of the modulus right? $\endgroup$ – lab bhattacharjee May 13 '17 at 9:23
  • $\begingroup$ Well, I wrote $\therefore \left | 1-cos\theta-isin\theta \right | = \sqrt{(1-cos\theta)^{2} + (-sin\theta)^{2}}$, then the square of the modulus would get rid of the square root, and then follows on to what I get in the end. I'm also a little confused with your response. :S $\endgroup$ – Skylineblue May 13 '17 at 9:26
  • $\begingroup$ @Skylineblue, How could you get rid of the square root? Do you mean $$\sqrt{a^2+b^2}=a^2+b^2$$ $\endgroup$ – lab bhattacharjee May 13 '17 at 9:29
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Just take $2$ common from the final result and apply the formulae $$1-\cos\theta=2\sin^2\frac{\theta}{2}$$ Your answer will be simplified.

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