Magnitude calculation of filter

Question

I am working on a problem for a given transfer function given below and am having trouble calculating the magnitude and phase of it. The problem is specifically that I do not know what method to use to seperate the imaginary value out of from the real to allow me to use their seperate values in calculating the magnitude and phase.
$$frac{Vout}{Vin}=frac{1}{1-(4*PI^2*0.04)+(j*2*PI*0.2)}$$
Ideally I'd like to end up with something like a+bi to allow me to do the srqt(a^2 + b^2) and the tan equation for the phase but I do not know how to do this.
Could anyone suggest a method to use or some mathematical identity to use?
Thanks!
So implementing what was in the first comment from ocrdu
It simplifies to
$$frac{Vout}{Vin}=frac{1}{(-0.579+j*1.25)}$$
Then multiplying top and bottom by inverse of bottom gives:
$$frac{Vout}{Vin}=frac{(0.579-j*1.25)}{(-0.579+j*1.25)*(0.579-j*1.25)}$$
But how does this help me?

smake5730 · Accepted Answer

So the solution was simply to take the square as follows:
$$frac{1^2}{sqrt{(1-(4*PI^2*0.04))^2+(j*2*PI*0.2)^2}}$$
j squared is j * j = -1
$$=frac{1^2}{sqrt{(1-(4*PI^2*0.04))^2+(-1*2*PI*0.2)^2}}$$
$$=0.722$$

ocrdu · Answer

$$frac{Vout}{Vin}=frac{-0.579-j*1.25}{(-0.579+j*1.25)*(-0.579-j*1.25)}=$$
$$frac{-0.579-j*1.25}{0.335+0.724j-0.724j+1.563} = $$
$$frac{-0.579-j*1.25}{1.898} = -0.305-0.659j$$
Please check for errors (yours and mine), but you get the idea.
Multiplying top and bottom by the denominator's complex conjugate gives a real number in the denominator.

Andy aka · Answer

$(1 + ja)cdot(1 - ja) = 1 + a^2$

Magnitude calculation of filter

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