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- #1

- Thread starter Markov
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- #1

- Jan 26, 2012

- 236

You can check that $e^{\pi i/4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$ and $e^{-\pi i/4} = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$.Solve ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{x}}=\sqrt{2}.$

Is there a faster way to solve this?

Therefore, the equation becomes,

$$ e^{ \pi i x/4} + e^{-\pi i x/4} = \sqrt{2} \implies \cos \left( \frac{\pi i x}{4} \right) = \frac{\sqrt{2}}{2} $$

Do you see how we got that?

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- #3

- Jan 26, 2012

- 236

Careful, those are not all of the solutions? Because $\cos -\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ also!Oh yes, you just used Euler's formula, so the solutions are $\dfrac{\pi }{4}ix = \dfrac{\pi }{4} \pm 2k\pi ,{\text{ }}k \in \mathbb{Z}$ and $\dfrac{\pi }{4}ix = \dfrac{{7\pi }}{4} \pm 2k\pi ,{\text{ }}k \in \mathbb{Z},$ are those correct?

Thanks for the help!

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- #5

I am a bit confused by the replies.Solve ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{x}}=\sqrt{2}.$

It clear that $x=\pm 1$ are solutions.

Was $x$ suppose to be complex also?

Moreover $\exp(i\theta)+\exp(-i\theta)=2\cos(\theta)$ not $2\cos(i\theta)$.

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- #7

Then might look at $x=\pm 7,~\pm 9,~\pm 23,~\pm 25,\cdots$.No, $x$ is supposed to be real, thanks for the catch Plato!

What is going on?

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