Question

How do you find `abs( 2-2i )`?

Answer 1

The answer is `=2sqrt2`

Explanation:

Let `z=a +ib` be a complex number.

Then the modulus of `z`is

`∣z∣=∣a+ib∣=sqrt(a^2+b^2)`

so,

`∣2-2i∣=sqrt(2^2+(-2)^2)`

`=sqrt(4+4)= sqrt8=2sqrt2`