# Complex Exp.

1. Aug 1, 2004

### kishtik

What is (a+bi)^(c+di) ? How can I find this?
=[(a+bi)^c]x[(a+bi)^di]=? Now I can go binomial for the first part but what about (a+bi)^di?

2. Aug 1, 2004

### pnaj

Have you dealt with exponentials and logarithms of complex numbers yet?

$$\begin{array}{l} \forall x,y \in {\rm R} \\ \exp (x + iy) = \exp (x)(\cos (y) + i\sin (y)) \\ \log _e (x + iy) = \log _e (\sqrt {x^2 + y^2 } ) + i\arctan (y/x) \\ \end{array}$$

(I've glossed over the fact that the log function is actually multi-valued ... let me know if you need this explained further).

$$\begin{array}{l} \forall z,w \in {\rm C }, w \neq 0 \\ \log _e (z^w ) = w\log _e (z) \\ z^w = \exp (w\log _e (z)) \\ \end{array}$$

See how you get on.

3. Aug 2, 2004

### kishtik

I couldn't understand the first three quations although I did the last three. And I have no idea about expotentials and logarithms of complex numbers. Thanks.

4. Aug 2, 2004

### sumeerbhatara

(a+ib)^{di} = ((m \exp (ni))^d)^i
m = \sqrt{a^2 + b^2}, n = arctan (b/a)
= (m \exp(ni))^i)^d
= ((m^i) \exp(-n))^d
= m^{id} \exp(-nd)
m^{id} is a complex number.

Regarding the equations,
They are pretty simple
The first equation is the famous De-Moivre's theorem. Prrof can be found in any algebra book.
For Eqn 2, from De-Moivre's theorem,
$$\begin{array}{l} x + iy = \sqrt(x^2+y^2)\exp(i\arctan(y/x)) \end{array}$$
Take logarithms on both sides and you get equation 2.

5. Aug 2, 2004

### pnaj

I'm not sure if there are ways to solve this without exp, log etc.

Does anyone know another way?

6. Aug 2, 2004

### Zurtex

Yeah, you really need to look up exponents, logs, trigonometric functions and hyperbolic functions in relationship to complex numbers otherwise you'll struggle to deal with such problems.