Medal - What's magnitude of just an imaginary number with no real part? I mean, I have iωL in a denominator...

Question

kainui · Answer

The magnitude of every imaginary number z is given by the square root of the number multiplied with it's complex conjugate.

$$|z| = \sqrt{zz^*}$$

It's clear to see why this is true cause of the pythagorean theorem:

$$z= a+bi$$$$z=a-bi$$

$$zz^* = (a+bi)(a-bi) = a^2+abi-abi+(i)(-i)b^2 = a^2 + b^2$$

Hopefully that helps

adamaero · Answer

Am I supposed to take it out when finding magnitude?
e.g., ||1/(iωL)|| =?= 1/(ωL)

adamaero · Answer

Am I missing something conceptually? 
All I think is i = $$\sqrt{-1}$$

kainui · Answer

You don't take it out, just follow the process, multiplying a number by its complex conjugate always gives a real number. Try it out

kainui · Answer

If you don't like that cause i is in the denominator then I can teach you a trick to juggle it up into the numerator:

Since $\sqrt{-1} = i$ we can just square it to get -1, nothing crazy:

$$i*i =-1$$
Now multiply both sides by -1
$$-i*i=1$$

Divide both sides by i
$$-i = \frac{1}{i}$$

So any time you see $\frac{1}{i}$ replace it with $-i$ since they're equal.

adamaero · Answer

but I can't have imaginary numbers as a part a magnitude