Engineering How to Calculate the Magnitude and Phase of the Transfer Function H(e^{jw})?

[jisbon]

Homework Statement

Find the magnitude and phase response of this system.
$H(e^{jw}) =\frac{1+e^{-jw}}{1-0.1e^{-jw}}$

Relevant Equations

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Tried this, but not sure how am I supposed to square the whole equation and then square root it since this will inevitably give me imaginary values. Am I supposed to ignore the imaginary values?
Also, how can I find out the phase in this case? Usually, it's taking the exponents but in this case, I'm not so sure what to do.

$H(e^{jw}) =\frac{1+e^{-jw}}{1-0.1e^{-jw}}$
$H(e^{jw}) =\frac{1+e^{-jw}}{1-0.1e^{-jw}} =\frac{1+cosw-isinw}{1+0.1cosw+0.1isinw}$
If I try to square it, and find the magnitude..

|$\frac{\left(-\sin ^2\left(w\right)+2\cos \left(w\right)+\cos ^2\left(w\right)+1\right)+\left(-2\sin \left(w\right)-\sin \left(2w\right)\right)i}{...}$|

Hence I'm stumped right here.. Anyone could point me in the correct direction? Thanks!

 

[Dr.D]

The transfer function you are starting with is a complex valued function. The magnitude is the square root of the sum of the squares of the real part and the imaginary part. Do you know how to do complex arithmetic? You have a complex number in the numerator and another in the denominator. Just apply the rules for complex algebra and it should all work out after quite a bit of scratch paper.

 

[boo]

Would not recommend you torture yourself by expanding the terms using Eulers formula until the very end. Instead simply multiply both numerator and denominator by the complex conjugate of the given denominator. The denominator will then reduce to a very simple number. You can now expand the numerator according to Euler. Of course you are going to get a complex number. It will have both a magnitude and a phase. At this point the magnitude is simply root the sum of the squares of real and imaginary parts and the angle will be the arctan of the ratio of imaginary to real parts.