Question

What does `(1-3i)/sqrt(1+3i)` equal?

Answer 1

`(1-3i)/sqrt(1+3i)`

`=(-2sqrt((sqrt(10)+1)/2)+3/2sqrt((sqrt(10)-1)/2))-(2sqrt((sqrt(10)-1)/2)+3/2sqrt((sqrt(10)+1)/2))i`

Explanation:

In general the square roots of `a+bi` are:

`+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)`

See: http://socratic.org/questions/how-do-you-find-the-square-root-of-an-imaginary-number-of-the-form-a-bi

In the case of `1+3i`, both Real and imaginary parts are positive, so it's in Q1 and has a well defined principal square root:

`sqrt(1+3i)`

`=sqrt((sqrt(1^2+3^2)+1)/2)+sqrt((sqrt(1^2+3^2)-1)/2)i`

`=sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i`

So:

`(1-3i)/sqrt(1+3i)`

`=((1-3i)sqrt(1+3i))/(1+3i)`

`=((1-3i)^2 sqrt(1+3i))/((1+3i)(1-3i))`

`=((1-3i)^2 sqrt(1+3i))/4`

`=1/4(1-3i)^2 (sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i)`

`=1/4(-8-6i)(sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i)`

`=-1/2(4+3i)(sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i)`

`=-1/2((4sqrt((sqrt(10)+1)/2)-3sqrt((sqrt(10)-1)/2))+(4sqrt((sqrt(10)-1)/2)+3sqrt((sqrt(10)+1)/2))i)`

`=(-2sqrt((sqrt(10)+1)/2)+3/2sqrt((sqrt(10)-1)/2))-(2sqrt((sqrt(10)-1)/2)+3/2sqrt((sqrt(10)+1)/2))i`