So, we know how to take the exponential of any complex number now. We do it by converting the exponential into the exponential of the real and imaginary parts separately, and then use the relationship between $e^{ia}$ and the $\cos$ and $\sin$ functions to write everything in terms of functions of real numbers, which we know how to deal with. How about the trigonometric functions applied to complex numbers? Well, we have a pretty good hint already from how we got from the exponential of complex numbers to trigonometric functions of real numbers. In fact we’re just going to give the answer, but you can work it out using Taylor series as well. For a complex number $z$:

$\cos z=\frac{e^{iz}+e^{-iz}}{2}$

$\sin z=\frac{e^{iz}-e^{-iz}}{2i}$

The first thing to check is that this is true when $z$ is a real number. It looks pretty strange at first site, especially the definition of $\sin$ because there’s an $i$ sticking out in the denominator like a sore thumb! How can this give us something real, as we know that it must? Let’s take $z=\theta$ where $\theta$ is real. Then:

$\cos \theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{\cos\theta+i\sin\theta+\cos(-\theta)+i\sin(-\theta)}{2}=\cos\theta$

$\sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}=\frac{\cos\theta+i\sin\theta-\cos(-\theta)-i\sin(-\theta)}{2i}=\sin\theta$

So that holds. How about trying to understand the behaviour of the trigonometric function of a purely imaginary complex number? Well, we know for sure that trig functions are always between -1 and 1, right? Well, it turns out that this is only true if $\theta$ is purely real. In fact if it’s complex then the trig functions can give you any complex number you like. We’ll see that shortly. Let $z=i b$ where $b$ is a real number, then:

$\cos ib=\frac{e^{-b}+e^{b}}{2}$

$\sin ib=\frac{e^{-b}-e^{b}}{2i}$

Then we can see that $\sin ib$ is purely imaginary and $\cos ib$ is purely real. The other thing that we can note is that for large positive or negative $b$, both of these numbers are going to be large in magnitude. In fact these functions have names $\cos ib=\cosh b$ which is known as hypergeometric cosine, and $-i\sin ib=\sinh b$ which is hypergeometric sine. We can plot both of these functions in the figure below and see that they don’t behave at all like the normal sine and cosine functions that we are used to.

There are hyperbolic trig identities just as there are for normal trig functions, but unlike the case where $\sin^2x+\cos^2x=1$, now $\cosh^2x-\sinh^2x=1$ In fact it is precisely this hyperbolic identity which gives the functions their name. Unlike the sine and cosine functions which describe circular behaviour, so the hyperbolic versions describe the behaviour of things moving along hyperboloids. In the following you can see the differences between hyperolic functions and circular functions, which are very close, up to a single sign in their definitions, but have completely different behaviours

OK, so let’s look at a very general example to see that actually we can have trig functions giving any result we want, not just numbers between -1 and 1 as we had when we were restricted to the real numbers. Let’s ask a very general question. Find $z$ when:

$\sin z=w$

For any complex number $w$. We start by writing the trig function in exponential form and then multiply both sides of the equation by $2i e^{iz}$. This gives:

$e^{2iz}-1=2wie^{iz}$

Now we note that we can write $e^{iz}=p$ and we can happily now write the equation as a normal quadratic equation:

$p^2-2 wip-1=0$

But we can solve this using the normal quadratic recipe (You can prove that this still holds, even for complex numbers, simply by using it and then plugging the result back into the quadratic equation). So:

$p=\frac{2wi\pm\sqrt{-4w^2+4}}{2}=wi\pm\sqrt{1-w^2}$

So, we now have that:

$e^{iz}=wi\pm\sqrt{1-w^2}$

How can we calculate what $z$ must be? Let’s start by saying that $z=a+bi$ so:

$e^{ia-b}=e^{-b}(\cos a+i\sin a)=wi\pm\sqrt{1-w^2}$

We can then solve this by taking the modulus and argument of both sides. First the modulus:

$|e^{-b}(\cos a+i\sin a)|=e^{-b}=|wi\pm\sqrt{1-w^2}|$

and the argument:

$arg\left(e^{-b}(\cos a+i\sin a)\right)=a=\arg\left(wi\pm\sqrt{1-w^2}\right)$

so now, given a $w$, we can calculate precisely the $z$ such that $\sin z=w$. You can see that you would be able to do this for any $w$ you are given, because any $w$ has both a modulus and an argument (though you’d have to be careful about $w=0$). to be a solution to the equation, thus, the full solution is:

$z=\arg\left(wi\pm\sqrt{1-w^2}\right)-i\ln|wi\pm\sqrt{1-w^2}|$

Remember that the argument of a complex number includes additions of $2n\pi$ with $n\in{\mathbb Z}$. This looks a bit abstract, but given specific values of $w$, we can calculate this relatively easily.

In the same vein as in the previous post, this is the mapping of lines in the complex plane under the Cosine function:

 How clear is this post?