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# Trig identities of the complex kind

I'm sure most of you have heard of trig identities involving $$\sin(x)$$, $$\cos(x)$$,$$\tan(x)$$, etc...

But what about complex trigonometry? Specifically the trig function $$\text{cis}(x)$$.

This function is short hand for $$\boxed{\cos(x) + i\sin(x)}$$ for those of you who haven't met it before. It's used mainly to represent and calculate complex numbers.

So what happens if you add two $$\text{cis}(x)$$ functions together? What about subtraction, multiplication, division, how does this all affect the $$\text{cis}(x)$$ function?

Before we get started, I'm going to list the identities I already know which involve this function. These and all future identities will be in boxes.

$\large \boxed{\text{cis}(x) = e^{xi}}$

$\large \boxed{(\text{cis}(x))^n = \text{cis}(nx)}$

Let's first expand on that second one. What if we replace $$n$$ with a complex number?

$(\text{cis}(x))^{a + bi}$

Knowing that $$\boxed{\text{cis}(x) = e^{xi}}$$ really helps here.

$(e^{xi})^{a + bi}$

$e^{(a + bi)xi}$

$e^{axi - bx}$

$\frac{e^{axi}}{e^{bx}}$

$\frac{\text{cis}(ax)}{e^{bx}}$

$\large \boxed{(\text{cis}(x))^{a + bi} = \frac{\text{cis}(ax)}{e^{bx}}}$

So now that that's updated we can go on to find new identities.

$\text{cis}(x)\text{cis}(y)$

$e^{xi}e^{yi}$

$e^{(x + y)i}$

$\text{cis}(x + y)$

$\large \boxed{\text{cis}(x)\text{cis}(y) = \text{cis}(x + y)}$

Now division.

$\frac{\text{cis}(x)}{\text{cis}(y)}$

$\frac{e^{xi}}{e^{yi}}$

$e^{xi}e^{-yi}$

$e^{(x - y)i}$

$\text{cis}(x - y)$

$\large \boxed{\frac{\text{cis}(x)}{\text{cis}(y)} = \text{cis}(x - y)}$

Alright so those are done now. Next is addition and subtraction.

$\text{cis}(x) + \text{cis}(y)$

$\text{cis}(x)\left(1 + \frac{\text{cis}(y)}{\text{cis}(x)}\right)$

$\text{cis}(x)(1 + \text{cis}(y - x))$

$\large \boxed{\text{cis}(x) + \text{cis}(y) = \text{cis}(x)(1 + \text{cis}(y - x))}$

And finally subtraction.

$\text{cis}(x) - \text{cis}(y)$

$\text{cis}(x)\left(1 - \frac{\text{cis}(y)}{\text{cis}(x)}\right)$

$\text{cis}(x)(1 - \text{cis}(y - x))$

$\large \boxed{\text{cis}(x) - \text{cis}(y) = \text{cis}(x)(1 - \text{cis}(y - x))}$

That's all for now, if there's any you think I've missed, please tell.

Edit: I forgot one: $$\text{arccis}(x)$$

$\text{cis}(x) = e^{xi}$

$\ln{\text{cis}(x)} = xi$

$\frac{\ln{\text{cis}(x)}}{i} = x$

$\frac{i\ln{\text{cis}(x)}}{-1} = x$

$-i\ln{\text{cis}(x)} = x$

$x = -i\ln{\text{cis}(x)}$

Replace $$x$$ with $$\text{arccis}(x)$$ and $$\text{cis}(x)$$ with $$x$$

$\large \boxed{\text{arccis}(x) = -i\ln{x}}$

Here's the new list of identities:

$\large \boxed{\text{cis}(x) = e^{xi}}$

$\large \boxed{(\text{cis}(x))^{a + bi} = \frac{\text{cis}(ax)}{e^{bx}}}$

$\large \boxed{\text{cis}(x)(\text{cis}(y))^{\pm 1} = \text{cis}(x \pm y)}$

$\large \boxed{\text{cis}(x) \pm \text{cis}(y) = \text{cis}(x)(1 \pm \text{cis}(y - x))}$

$\large \boxed{\text{arccis}(x) = -i\ln{x}}$

Note by Jack Rawlin
6 months, 3 weeks ago

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