Why is |sqrt(i)| = 1? (Imaginary Unit Explanation)

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Why is |sqrt(i)| = 1? (where i is the imaginary unit)

I thought to find a norm you do sqrt(x^2 + y^2) so isn't it
sqrt(0 + sqrt(i)^2)
= sqrt(i)

??

What am I doing wrong here?

nvm, i got it...
 
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For anyone else reading this and curious about a similar issue, it is important that you recognize that y in the formula for the norm is simply the imaginary part of the complex number (which is real, of course).

Of course, the other way to calculate the norm is to explicitly find i^{1/2} and find the norm of this. You will find that it is the same.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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