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.

We'll start with multiplication.

\[\text{cis}(x)\text{cis}(y)\]

\[e^{xi}e^{yi}\]

\[e^{(x + y)i}\]

\[\text{cis}(x + y)\]

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

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:

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

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

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

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

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

## Comments

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TopNewestThis is a great introduction to complex trigonometric functions. Can you add that to the page? – Calvin Lin Staff · 10 months, 3 weeks ago

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