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I came across this:

$$\frac{1}{e^{ib}} = \frac{2i}{(\sqrt{2}i+a)}$$

And I was wondering if there is a real solution for $b$ and $a$.

My attempt is solving for $a$

$$a+\sqrt{2}i= 2ie^{ib}$$

$$a= 2ie^{ib}-\sqrt{2}i$$

But the only real solution would be if $a=0$ and $e^{ib}=\sqrt{2}$ but this is impossible.

When I solve for $b$ I end up with this:

$$\frac{a+\sqrt{2}i}{2i}= e^{ib}$$

Here I'm not sure how to handle this, since I believe I can not just take the logarithm. Also Euler's formula doesn't help. This doesn't have a real solution too, right?

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    If you're hoping that $a$ and $b$ are real, then the right hand side has modulus 1. So the left hand side has modulus 1. What's that tell you about $a$? – 2017-01-27

3 Answers 3

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Take the absolute value of both sides and you will find that we need

$$|a+i\sqrt2|=2$$

or,

$$\sqrt{a^2+2}=2$$

$$\implies a=\pm\sqrt2$$

And likewise, substituting $a=+\sqrt2$ and solving for $b$:

$$e^{-bi}=\frac{i\sqrt2}{1+i}=\frac{i\sqrt2(1-i)}{(1+i)(1-i)}=\frac{1+i}{\sqrt2}$$

This is at an angle of $\pi/4$, thus, $(a,b)=(\sqrt2,-\pi/4)$.

Likewise, the other solution is given by $(-\sqrt2,\pi/4)$.

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    Thank you very much. I was not thinking of the absolute value. – 2017-01-27
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    @MarcE Your welcome :-) – 2017-01-27
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Hint: for $a,b \in \mathbb{R}\,$:

$$e^{ib} = \frac{\sqrt{2}i+a}{2i} \iff \cos b + i \sin b = \frac{1}{\sqrt{2}} - \frac{a}{2}i \iff \cos b = \frac{1}{\sqrt{2}} \;\;\text{and}\;\; \sin b = - \frac{a}{2}$$

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    Perfectly simple +1 – 2017-01-27
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    @SimplyBeautifulArt Thanks. I just couldn't let this pass: "*Also Euler's formula doesn't help*" ;-) – 2017-01-27
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    Thank you, yes very simple. – 2017-01-27
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Hint: If b is a solution then $b+2\pi k$ is also a solution. it is easy to see that $b=1/i\ln(\sqrt{2})$ solves the equation.

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    Thank you also for the hint :) – 2017-01-27