Trouble with Complex Exponentials

[NewtonianAlch]

Homework Statement

Show that if Im(z) ≥ 0, then |${{\rm e}^{i \left( x+iy \right) }}$| ≤ 1.

Let R > 1 be a real constant

Now deduce that ${\frac {{{\rm e}^{-{\it Im} \left( x+iy \right) }}}{ \left| \left( x+<br /> iy \right) ^{4}+1 \right| }}$ ≤ $\left( {R}^{4}-1 \right) ^{-1}$ for z on the semi-circle {z $\in$ ℂ: |z| = R, Im(z) ≥ 0}

The Attempt at a Solution

The imaginary part of z ≥ 0 is just y ≥ 0.

Multiplying the exponential through, you get, ${{\rm e}^{ix-y}}$.

I realize that taking the modulus of this will make the e^ix disappear, leaving only e^-y, which clearly shows then as y ≥ 0:

e^-y ≤ 1

I'm not too sure how to get the modulus of that exponential.

N.B: The modulus applies to the entire fraction, I cannot seem to get the modulus including the entire fraction for some reason, it's just appearing on the denominator

 

[NewtonianAlch]

I tried a few methods, with this I got what looked correct, but I wasn't really sure of what I was doing.

${{\rm e}^{ix}}{{\rm e}^{-y}}$

${{\rm e}^{-y}} \left( \cos \left( x \right) +i\sin \left( x \right) <br /> \right)$

Taking the absolute value of that:

Square Root of $[\left( {{\rm e}^{-y}} \right) ^{2} \left( \cos \left( x \right) <br /> \right) ^{2}+ \left( {{\rm e}^{-y}} \right) ^{2} \left( \sin \left( x<br /> \right) \right) ^{2}]$

Square Root of $[{{\rm e}^{-2\,y}}*(1)]$

Which is ${{\rm e}^{-y}}<br />$

Voila? So now we can see that as y increases the value of e^-y is going to decrease, with its maximum value when y = 0 as per I am (z).

Edit: Forgot to add the second part of the question earlier, just added it in, not sure on how to proceed with that part.