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Complex equation

Markov

Member
Feb 1, 2012
149
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?
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
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?
You can check that $e^{\pi i/4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$ and $e^{-\pi i/4} = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}$.

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?
 

Markov

Member
Feb 1, 2012
149
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!
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
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!
Careful, those are not all of the solutions? Because $\cos -\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ also!
 

Markov

Member
Feb 1, 2012
149
Oh yes! So the we have these sets also: $ - \displaystyle\frac{\pi }{4}ix = \frac{\pi }{4} \pm 2k\pi ,{\text{ }}k \in \mathbb{Z},{\text{ }} - \frac{\pi }{4}ix = \frac{{7\pi }}{4} \pm 2k\pi ,{\text{ }}k \in \mathbb{Z},$
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Solve ${{\left( \dfrac{1+i}{\sqrt{2}} \right)}^{x}}+{{\left( \dfrac{1-i}{\sqrt{2}} \right)}^{x}}=\sqrt{2}.$
I am a bit confused by the replies.
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 $2\cos(i\theta)$.
 

Markov

Member
Feb 1, 2012
149
No, $x$ is supposed to be real, thanks for the catch Plato!
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
No, $x$ is supposed to be real, thanks for the catch Plato!
Then might look at $x=\pm 7,~\pm 9,~\pm 23,~\pm 25,\cdots$.
What is going on?
 

Markov

Member
Feb 1, 2012
149
Mmm you mean would have to rephrase the condition for $x$ ?
 

Markov

Member
Feb 1, 2012
149
What are the solutions then? Do we have to reformule the problem?