# How to prove that $j^\mu=\cos(\frac{\mu\pi}{2})+j\sin(\frac{\mu\pi}{2})$? [closed]

I am trying to read a journal paper and I need to understand some math operations.

At this time, I feel it hard to understand this equation:

$j^\mu=\cos(\frac{\mu\pi}{2})+j\sin(\frac{\mu\pi}{2}), j=sqrt(-1), \mu\in R,$

Is this something from the complex number angular formation definition or some kind of Euler equation? Last year, I have asked my professor to help me go through this part. But now I still can't feel familiar to this part.

Based on some answers:

$j= e^{j \frac{\pi}{2}}$ is the part that I feel hard to understand.

I know some basic complex number operations and I often use MATLAB to solve math problem.

Thanks,

Bo

## closed as unclear what you're asking by Did, Sahiba Arora, Namaste, Daniel W. Farlow, Claude LeiboviciJul 30 '17 at 4:32

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• How do you define complex exponentiation? Might be worth looking up beforehand – Wojowu Jul 29 '17 at 22:09
• If $\mu\in\mathbb{Z}$ then this is from the De Moivre Formula... and you can prove it by induction. :) – Ixion Jul 29 '17 at 22:10
• @Wojowu It is really a good idea to look up the definition of complex exponentiation. – Bo Shang Jul 29 '17 at 22:24
• "It is really a good idea to look up the definition of complex exponentiation" ?? Thus... did you look it up? – Did Jul 29 '17 at 22:55
• It's more complicated, need understanding of holomorphic functions: en.wikipedia.org/wiki/Complex_logarithm Perhaps you haven't red about this before? – Lehs Jul 29 '17 at 22:59

Updating my answer, as OP has said that $\mu\in \mathbb R$, so the assumption I made in my previous answers below that $\mu \in \mathbb Z$ no longer hold.

The first question to ask is, "What does $j^{\mu}$ even mean if $\mu$ is not an integer?" There are three numbers that satisfy the equation, $x^3=j$, and while you could argue that one of them should be the value $j^{1/3}$, which of the three values should you take?

One way around this is by appealing to the exponential function, $\operatorname{exp}(x)=e^x=\sum \frac{x^n}{n!},$ where the sum is our actual definition, and $e^x$ is suggestive shorthand. One can establish various properties like $e^{a+b}=e^a e^b$, which mean that $e^x$ behaves like normal exponentiation in cases where things are simple, i.e., if $e^x=\alpha$, then $e^{nx}=\alpha^n$, and $\left(e^{x/n}\right)^n=\alpha$. Since $e^x>0$ whenever $x\in \mathbb R$, this allows us to define $x^{1/n}$ as $e^{\log x/n}$ for positive $x$, which always spits out the real $n$th root instead of one of the other $n-1$ complex roots we could have taken. It also allows us to define $x^{\mu}=e^{\mu\log x}$ for positive $x$ even when $\mu$ is not rational. Because of continuity of $e^x$, and we have that if $(\r_i)$ is a sequence of rational numbers approaching $\mu$, then by taking the exponential function to define $x^{r_i}$, we have $x^{r_i}\to x^{\mu}$, and so this definition fits in with just saying "define things as roots of polynomials, then extend using continuity."

However, we run into problems defining $x^{\mu}$ when $x$ is not a positive real number. I don't want to get into all the ways that things can go wrong, but what I am about to do is not wholly sufficient.

By using the sum formula, we can define $e^x$ not just for $x\in \mathbb R$, but actually for $x\in \mathbb C$. By looking at the Taylor series for $\sin$ and $\cos$ (or otherwise), we can establish that $e^{jx}=\cos x + j\sin x$, and we can check that $e^{a+b}=e^a e^b$ still holds when $a$ and $b$ are complex numbers.

Now, we can answer your question. Since $e^{j*\pi/2}=j$, we can define $j^{\mu}=e^{j\pi \mu/2}=\cos(\pi\mu/2)+j\sin(\pi \mu/2)$, and this will agree with the values you would get for $j^{\mu}$ when $\mu$ is an integer. However, this is not the only way to go...

Since $e^{2j\pi}=1$, we also have that $e^{j*\pi/2+2jk\pi}=j$ whenever $k$ is an integer, and so we could define $j^{\mu}=e^{j\mu(\pi/2+2k\pi)}$ for some integer $k$. This will give us the same values when $\mu$ is an integer, but will give different values when $\mu$ is not. While one can argue that $k=0$ gives the simplest possibility, there isn't a compelling argument (that I can think of) to say that it is the "right" definition in any real sense.

Just to throw something interesting in, the above argument shows that there are an infinite number of possibilities for what the value $j^j$ should be. However, all of them are real numbers!

You can prove the formula by induction, but for the induction step you are going to need the following result, which is easy enough to prove by expanding things out, regrouping terms, and making an appeal to standard trig identities:

$$\left(\cos(\theta)+j\sin(\theta)\right)\left(\cos(\varphi)+j\sin(\varphi) \right)=\cos(\theta+\varphi)+j\sin(\theta+\varphi)$$

A second approach that you could take (because you are only asking about $j^{\mu}$ and not powers of a different complex number) is to verify that both the left and right hand sides of your equation only depend on the remainder of $\mu$ when divided by $4$, and then break things up into four cases, depending on if that remainder is $0, 1, 2$, or $3$.

• by induction The question doesn't specify that $\mu$ is an integer (though that's likely what the OP means). – dxiv Jul 29 '17 at 22:31
• Calculating $a^b$ if $a$ isn't a positive real number and $b$ isn't an integer isn't really well defined. I don't feel bad about giving an answer that only holds in the case where one doesn't need to struggle to figure out what the problem actually means. – Aaron Jul 29 '17 at 22:37
• The OP has since clarified that $\,\mu \in \mathbb{R}\,$. I suggest you at least preface your answer with assuming μ is a positive integer .... – dxiv Jul 29 '17 at 23:18
• Interesting addendum! – Ixion Jul 30 '17 at 1:58

Please, check my answer, I'm new in Complex Analisys, but I try my best.

$$j=0+1\cdot j\implies j=\rho e^{j\theta}$$ where $\rho=\sqrt{[\mbox{Re}(j)]^2+[\mbox{Im(j)}]^2}=1 \ \mbox{and} \ \theta=\arg(j)=\frac{\pi}{2}$. So $j$ in exponential form is: $j=e^{j\frac{\pi}{2}}$ According with the property of exponential: $$(e^a)^b=e^{(a+2k\pi j)\cdot b} \ \ \ \mbox{with} \ a,b\in\mathbb{C}\wedge k\in\mathbb{Z}$$ we have: $$j^{\mu}=(e^{j\frac{\pi}{2}})^{\mu}=e^{(j\frac{\pi}{2}+2k\pi j)\cdot \mu}$$

If $\mu\in\mathbb{Z}$ then $e^{(j\frac{\pi}{2}+2k\pi j)\cdot \mu}$ becomes $$e^{(j\frac{\pi}{2}+2k\pi j)\cdot \mu}=\cos\left(\frac{\pi}{2}\mu+2k\mu\pi\right)+j\sin\left(\frac{\pi}{2}\mu+2k\mu\pi\right)=\cos\left(\frac{\pi}{2}\mu\right)+j\sin\left(\frac{\pi}{2}\mu\right)$$ because $\mu\cdot k\in\mathbb{Z}$ hence we can use the periodicity of sine and cosine functions. If $\mu\in\mathbb{R}-\mathbb{Z}$, than $k\cdot\mu$ may not be an integer hence we can not use periodicity. The only thing we can write is: $$e^{(j\frac{\pi}{2}+2k\pi j)\cdot \mu}=\cos\left(\frac{\pi}{2}\mu+2k\mu\pi\right)+j\sin\left(\frac{\pi}{2}\mu+2k\mu\pi\right)$$ These problems arise from the periodicity of exponential function.

$j^μ=e^{jπμ/2}=\cos (πμ/2)+j\sin (πμ/2)$

• Thank you for your answer. @Blumer I just looked up my notes from last year. It is like what you mentioned. However, $j^\mu=e^(j\pi\mu/2)$ is the part that I do not really understand. – Bo Shang Jul 29 '17 at 22:28
• One-line posts rarely make good answers, particularly when they are in larger part a repetition of something posted in the Question. Perhaps you can address the concern expressed in the OP's Comment above? – hardmath Jul 29 '17 at 22:37
• All right sorry – Blumer Jul 30 '17 at 8:09