# Proving $|e^{iθ}|=1$

How do I show that $|e^{iθ}|=1$? So I got that the length will be $\sqrt{\cos^2(x)-\sin^2(x)}$ and it can be written as the square root of $\cos 2x$ but I don't see how that equals 1.

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$|a+bi|=\sqrt{a^2+b^2}$ not $\sqrt{a^2+(ib)^2}$ –  anon Sep 6 '13 at 21:47
When $x= \pi$ your formula yields $\sqrt{-1}$ as the length, that should be a sign that the formula you are using is not the right one ;) –  N. S. Sep 7 '13 at 16:33

For a complex number of the form $z=a+ib$, it's magnitude is given by $|z|=\sqrt{a^2+b^2}$.

I believe you have incorrectly used $|z|=\sqrt{a^2-b^2}$

See here: Magnitude of a Complex Number

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The length will be $\sqrt{\cos^2(\theta )+\sin^2(\theta)}=1$.

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According to de Euler's formula you have that $$e^{i\theta} = \cos(\theta) + i \sin (\theta)$$ Now the module $| \cdot |$ of a complex number is $x+i y$ is simply $(x^2+y^2)^{1/2}$. Note that we are ignoring the $i$ when taking squares. I think you derived your incorrect formula because you also squared the $i$. Use also that for any real number $\theta$ we have: $\sin^2(\theta)+\cos^2(\theta)=1$.

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From $e^z = \sum_{k = 0}^{\infty} \frac{z^k}{k!}$, we see that $\overline{e^z} = e^{\overline{z}}$, and from $e^a e^b = e^{a+b}$ we get

$$|e^z|^2 = e^z \times\overline{e^z} = e^z e^{\overline{z}} = e^{z + \overline{z}}$$

In particular, for $z = i\theta$, you get $|e^{i\theta}| = 1$.

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