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Given this complex number,

$$e^{9ix/2} \frac{\sin 4x }{ (\sin (x/2) }$$

The real part of this complex number can be worked out easily, by replacing the $e^{9ix/2}$ with $\cos(9x/2)$

However if I'm given the complex number,

$$\frac{3} {3 - e^{ix} }$$

I cannot work out the real part by replacing the $e^{ix}$ with $\cos x $.

I want to understand why I can do this replacement in the first example and why I can't do it in the second example; and what I should look for when doing practice questions myself. And how I would actually go about working out the real part of the second example.

Thanks, any help would be appreciated.

Trunkersn
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2 Answers2

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Replacement is a good strategy for both, but you have to do additional steps for the second problem.

Note that the first one is of the form $$ (a + ib)c $$ where $a,b$ and $c$ are all real numbers, so you just have to do one multiplication to get a result of the form $x + iy$ from which it is easy to determine the real (or imaginary) part.

In the second case, you should still go ahead and expand the exponential, you'll have something of the form $$ \frac{a}{b + ic}. $$ This cannot be easily written as $x + iy$, but try multiplying by $\frac{b-ic}{b-ic}$ and see if that helps.

BaronVT
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  • Thanks for the reply. Doing this I got the correct answer, but I still failed to see why we need to do this, both yours and @gimusi solutions although different still achieve the same thing which is getting rid of the complex number in the denominator and then we replace. Why can't we just replace in the denominator why do we need to do this? So if the complex number in the numerator was "(a+ib) + c", the real part would be a+c (over the denominator), but if this same thing was in the denominator we cannot do this? Why is this? Thanks. – Trunkersn Oct 24 '18 at 11:21
  • division distributes over addition in the numerator, but not the denominator.

    in other words, it is valid to write $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$, but on the other hand it is not generally true that $\frac{a}{b+c} = \frac{a}{b} + \frac{a}{c}$.

    Thus, if you have a complex number (real and imaginary part) in your denominator, you will have to do some work before you can split it into real and imaginary parts, whereas the case where the complex number in the numerator is more straightforward.

     – BaronVT Oct 24 '18 at 20:15
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    Thank you very much, that cleared everything up. – Trunkersn Oct 25 '18 at 09:47
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HINT

Use that

$$\frac{3} {3 - e^{ix} }=\frac{3} {3 - e^{ix} }\frac{3 - e^{-ix} } {3 - e^{-ix} }=\frac{9 - 3e^{-ix} } {10 -3 (e^{ix}+e^{-ix}) }$$

then recall that $z+\bar z= 2\Re(z)$.

user
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