15. Complex trigonometric functions
ok, so now you can calculate where in the complex plane a number like $e^{-2+3i}$ is by writing this in terms of trig functions:
$$e^{-2+3i}=e^{-2}\left(\cos 3+i\sin 3\right)$$.
and all the stuff on the right you can calculate very easily on even the simplest of calculators.
How about $\cos(-2+3i)$. What is this? Where does it sit in the complex plane? Well, you certainly can't type this into a basic calculator. How can we work out trig functions applied to complex numbers? Well, we're going to use some clever footwork and take advantage of what we did in the previous steps. Here it is:
Recall Euler's formula:
$$ e^{i\theta}=\cos{\theta}+i\sin{\theta} \qquad\qquad(1)$$
Let $\theta =-\theta$:
$$e^{-i\theta}=\cos{\theta}-i\sin{\theta} \qquad\qquad(2)$$
(1)+(2) yields:
$$2\cos{\theta}=e^{i\theta}+e^{-i\theta} \Rightarrow \cos{\theta}=\dfrac{e^{i\theta}+e^{-i\theta}}{2}$$
(1)-(2) yields:
$$\sin{\theta}=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$$
See if you can obtain the above formulae using the properties $z+\overline{z}=2Re(z)$ and $z-\overline{z}=2iIm(z)$ (Hint: choose $z=e^{i\theta}$).
We use the equations above as a motivation for the definitions that follow.
The complex trigonometric function
Given any complex number $z$, we define :
$\sin{z}=\dfrac{e^{iz}-e^{-iz}}{2i}$ and $\cos{z}=\dfrac{e^{iz}+e^{-iz}}{2}$
Remark: One big difference between real and complex trigonometric functions is that real sine and cosine functions are bounded between -1 and 1, while the complex functions are not bounded if they are defined over the set of all complex numbers (see examples that follow).
So, how could we use this to write $\cos(-2+3i)$ in Cartesian form?
Well, according to the above definition:
$$\cos(-2+3i)=\frac{e^{i(-2+3i)}+e^{-i(-2+3i)}}{2}=\frac{e^{(-3-2i)}+e^{(3+2i)}}{2}$$
And now we can use the definition of the complex exponential to give us:
$$\cos(-2+3i)=\frac{e^{-3}(\cos(-2)+i\sin(-2))+e^{3}(\cos(2)+i\sin(2))}{2}=\frac{e^{-3}\cos (-2)+e^3\cos 2}{2}+i\frac{e^{-3}\sin(-2)+e^3\sin 2}{2}$$
which can be written as:
$$\cos(-2+3i)=\frac{(e^{-3}+e^3)\cos 2}{2}+i\frac{(e^3-e^{-3})\sin 2}{2}$$
and all of this can be calculated on a basic calculator. It looked like quite a lot of work, but the steps are all pretty simple. This means that you can now calculate the $\sin$ and $\cos$ of any complex number you want and know where it sits in the complex plane.
This is not going to be the only time you will see "trigonometric functions" related to exponential functions. You'll come across (if you haven't already) hyperbolic functions whose definitions resemble those of the complex trigonometric functions. The most common hyperbolic functions are the hyperbolic sine, denoted $\sinh$, and the hyperbolic cosine, denoted $\cosh$, defined by:
$$\sinh{x}= \dfrac{e^x-e^{-x}}{2}$$ and $$\cosh{x}=\dfrac{e^x+e^{-x}}{2} $$
Hyperbolic functions may not be periodic functions as are the trigonometric functions; but many identities involving hyperbolic functions are quite similar to those for the corresponding trigonometric functions, as suggested by the names.
The following identities remain valid and can be proved using definition 5.2 and properties of the exponential function. Proofs are left to the Check Your Understanding exercises.
Properties of the complex trigonometric functions
If $z$ is a complex number, then:
- $\sin{(z+2\pi)}=\sin{z}$, and $\cos{(z+2\pi)}=\cos{z}$
- $\sin{(-z)}=-\sin{z}$ , and $\cos{(-z)}=\cos{z}$
- $\sin{2z}=2\sin{z}\cos{z}$ , and $\cos{2z}=\cos^2{z}-\sin^2{z}$
- $\sin^2{z}+\cos^2{z}=1$
Remark: Property 15.1 (a) tells us that $\sin{z}$ and $\cos{z}$ are both periodic with period $2\pi$. This also tells us that $\sin{z}$ and $\cos{z}$ are not one-to-one functions, preventing them from have inverse functions.
Find complex number(s) $z$ that satisfy the following equations:
- $\cos{z}=2$
- $\sin{z}=2$
Show solutions
The method for these is always the same...we take the following steps:
- Write the trig function in terms of exponentials: $$\frac{e^{iz}+e^{-iz}}{2}=2$$
- Multiply everything through by 2 in this case, and always by $e^{iz}$: $$e^{2iz}+1=4e^{iz}$$
- Let $w=e^{iz}$ so that we have a quadratic in $w$: $$w^2-4w+1=0$$
- Complete the square: $$(w-2)^2-3=0$$
- Solve the quadratic: $$w=2\pm\sqrt{3}$$
- Check that both of these solutions are positive (we have to do something slightly different if they're not.)
- Now replace the $w$ with $e^{iz}$. Let $z=a+bi\implies e^{iz}=e^{-b}e^{ia}$, and write the right hand side in complex exponential form: $$e^{-b}e^{ia}=(2\pm \sqrt{3})e^{2\pi i k}\, ,\,\,\, k\in\mathbb{Z}$$
- Match the modulus and argument of both sides: $$e^{-b}=2\pm \sqrt{3}\implies b=-\ln(2\pm\sqrt{3})\, ,\,\,\, a=2\pi k$$
- Write $z$ in terms of these $a$ and $b$ that you've just found: $$z=2\pi k-i\ln(2\pm\sqrt{3})\, , \,\,\, k\in\mathbb{Z}$$
Very similar to the last, but we'll see a subtlety:
- Write the trig function in terms of exponentials: $$\frac{e^{iz}-e^{-iz}}{2i}=2$$
- Multiply everything through by $2i$ in this case, and always by $e^{iz}$: $$e^{2iz}-1=4ie^{iz}$$
- Let $w=e^{iz}$ so that we have a quadratic in $w$: $$w^2-4iw-1=0$$
- Complete the square: $$(w-2i)^2+3=0$$
- Solve the quadratic: $$w=2i\pm\sqrt{3}i=i(2\pm\sqrt{3})$$
- Here are two complex roots. We have to write them in modulus argument form. Their modulus is $2\pm\sqrt{3}$ and all of their arguments are of the form $\frac{\pi}{2}+2\pi k$, $k\in\mathbb{Z}$. So again writing everything back in terms of $z=a+bi$ we have: $$e^{-b}e^{ia}=(2\pm\sqrt{3})e^{\frac{\pi}{2} i+2\pi k i}$$ Matching moduli and arguments we have $$z=\frac{\pi}{2}+2\pi k-i\ln(2\pm\sqrt{3})$$
Solve for $z \in \mathbb{C}$ if:
- $\sin{z}=4i$
- $\cos{2z}=2$
- $\cos{z}=\sin{z}$
- $\sin{(iz)}=\dfrac{e^z}{2}$
- Show that the formula $e^{iz}=\cos{z}+i\sin{z}$ holds true for all complex numbers $z$.
- Prove that $\sin{z}=0$ if, and only if, $z=k\pi$, where $k$ is an integer.
- Prove that $\cos{z}=0$ if, and only if, $z=\frac{\pi}{2}+k\pi$, where $k$ is an integer.
- Prove Properties 15.1.