Trouble with Complex Exponentials

  • #1

Homework Statement


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

Let R > 1 be a real constant

Now deduce that [itex]{\frac {{{\rm e}^{-{\it Im} \left( x+iy \right) }}}{ \left| \left( x+
iy \right) ^{4}+1 \right| }}
[/itex] ≤ [itex] \left( {R}^{4}-1 \right) ^{-1}[/itex] for z on the semi-circle {z [itex]\in[/itex] 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, [itex]{{\rm e}^{ix-y}}[/itex].

I realise 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
 
Last edited:

Answers and Replies

  • #2
I tried a few methods, with this I got what looked correct, but I wasn't really sure of what I was doing.

[itex]{{\rm e}^{ix}}{{\rm e}^{-y}}[/itex]

[itex]
{{\rm e}^{-y}} \left( \cos \left( x \right) +i\sin \left( x \right)
\right)
[/itex]

Taking the absolute value of that:

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

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

Which is [itex] {{\rm e}^{-y}}

[/itex]

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 Im (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.
 
Last edited:

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