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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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    $\begingroup$ $|a+bi|=\sqrt{a^2+b^2}$ not $\sqrt{a^2+(ib)^2}$ $\endgroup$
    – anon
    Sep 6, 2013 at 21:47
  • $\begingroup$ 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 ;) $\endgroup$
    – N. S.
    Sep 7, 2013 at 16:33

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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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Even if you wouldn't use the distance formula, when you graph z = cost + isint you will see in the triangle that all points Z are on the unit circle? That's really fundamental trigonometry. The hypotenuse is traditionally 1. This is the unit circle approach in the introduction of trigonometry and that comes back when trig finds its place as a representation of complex numbers in polar form

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