Finding the length of a complex vector

Question

Let us say the vector $u = (1 - i, 2 + 3i, 5)$.

How would I find $\|u\|$?

I have tried solving this through:

$\sqrt{(1-i)^2 + (2+3i)^2 + 5^2}$

But I ended up needing to find the square root of a complex number, which from what I understand is impossible.

That’s not the definition of the norm of a complex vector. The correct one is $|u|=\sqrt{u\overline u}$. — amd, May 16 '17 at 20:13
Is that the same as taking the complex conjugate of each item in the vector? I'm only aware of complex conjugates for complex numbers. — Daniel Paczuski Bak, May 16 '17 at 20:26

score 4 · Answer 1 · answered May 16 '17 at 20:18

For a complex vector $z = (z_1, \cdots, z_n)$, its norm is defined by

$$ ||z|| = \sqrt{\sum_{i=1}^n z_i\bar{z_i}}.$$

So each term $z_i \bar{z_i}$ in the sum is real and the resulting sum is real. In your case, $||u|| = \sqrt{(1-i)(1+i) + (2+3i)(2-3i) + (5)(5)} = \sqrt{2 + 13 + 25} = 2 \sqrt{10}.$

Sidenote: It is possible to take square roots of complex numbers. To see why, consider the geometric interpretation. A complex number at $(r,\theta)$ in polar coordinates will have the square roots $(\sqrt{r}, \theta/2)$ and $(-\sqrt{r}, \theta/2)$. This is because the angles add and the magnitudes multiply during complex multiplication. For example,

$$ \sqrt{i} = \pm (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i ).$$

It is more common to use $(\sqrt r, \theta/2+\pi)$ rather than $(-\sqrt r, \theta/2)$. — Arthur, May 16 '17 at 22:42

Finding the length of a complex vector

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