Complex Exponentials and Trig Identities

Recall that

$$r_1(\cos(\theta_1) + i \sin(\theta_1)) r_2(\cos(\theta_2) + i \si... ...eta_2)) = (r_1 r_2) (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)).$$

The angles add. You've seen something similar before:

$$e^a e^b = a^{a+b}.$$

This connection between exponentiation and (4.4) gives us an idea!

If $z=x+iy$ is a complex number, define

$$e^z = e^x(\cos(y) + i\sin(y)).$$

We have just written polar coordinates in another form. It's a shorthand for the polar form of a complex number:

$$r(\cos(\theta) + i\sin(\theta)) = re^{i\theta}.$$

Theorem 4.4.1   If $z_1$, $z_2$ are two complex numbers, then

$$e^{z_1} e^{z_2} = e^{z_1+z_2}$$

Proof.

$$e^{z_1} e^{z_2} = e^{a_1}(\cos(b_1) + i\sin(b_1)) \cdot e^{a_2}(\cos(b_2) + i\sin(b_2))$$

$$= e^{a_1+a_2}(\cos(b_1+b_2) + i\sin(b_1 + b_2))$$

$$= e^{z_1+z_2}.$$

Here we have just used (4.4). $\qedsymbol$

The following theorem is amazing, since it involves calculus.

Theorem 4.4.2   If $w$ is a complex number, then

$$\frac{d}{dx} e^{w x} = w e^{wx},$$

for $x$ real. In fact, this is even true for $x$ a complex variable (but we haven't defined differentiation for complex variables yet).

Proof. Write $w=a+bi$.

$$\frac{d}{dx} e^{w x} = \frac{d}{dx} e^{ax +bix}$$

$$= \frac{d}{dx} (e^{ax} (\cos(bx) + i \sin(bx)))$$

$$= \frac{d}{dx} (e^{ax} \cos(bx) + i e^{ax}\sin(bx))$$

$$= \frac{d}{dx} (e^{ax} \cos(bx)) + i \frac{d}{dx} (e^{ax}\sin(bx))$$

Now we use the product rule to get

$$\frac{d}{dx} (e^{ax} \cos(bx)) + i \frac{d}{dx} (e^{ax}\sin(bx))$$

$$= ae^{ax} \cos(bx) - be^{ax}\sin(bx) + i (a e^{ax}\sin(bx) + b e^{ax}\cos(bx))$$

$$= e^{ax}(a \cos(bx) - b\sin(bx) + i (a \sin(bx) + b \cos(bx))$$

On the other hand,

$$w e^{wx} = (a+bi) e^{ax + bxi}$$

$$=(a+bi) e^{ax} (\cos(bx) + i \sin(bx))$$

$$=e^{ax}(a+bi)(\cos(bx) + i \sin(bx))$$

$$= e^{ax}((a\cos(bx) - b\sin(bx)) + i ( a\sin(bx)) + b\cos(bx))$$

Wow!! We did it! $\qedsymbol$

That Theorem 4.4.2 is true is pretty amazing. It's what really gets complex analysis going.

Example 4.4.3   Here's another fun fact: $e^{i\pi} + 1=0.$
Solution. By definition, have $e^{i\pi} = \cos(\pi) + i \sin(\pi) = -1 + i 0 = -1$.