Trig identities of the complex kind

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

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

This function is short hand for cos(x)+isin(x)\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 cis(x)\text{cis}(x) functions together? What about subtraction, multiplication, division, how does this all affect the cis(x)\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.

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

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

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

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

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

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

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

eaxibxe^{axi - bx}

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

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

(cis(x))a+bi=cis(ax)ebx\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.

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

exieyie^{xi}e^{yi}

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

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

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

Now division.

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

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

exieyie^{xi}e^{-yi}

e(xy)ie^{(x - y)i}

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

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

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

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

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

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

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

And finally subtraction.

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

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

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

cis(x)cis(y)=cis(x)(1cis(yx))\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: arccis(x)\text{arccis}(x)

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

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

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

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

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

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

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

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


Here's the new list of identities:

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

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

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

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

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

Note by Jack Rawlin
3 years, 7 months ago

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This is a great introduction to complex trigonometric functions. Can you add that to the page?

Calvin Lin Staff - 3 years, 7 months ago

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