Calculating (a+bi)^(c+di): How to Find it?

[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?

 

[pnaj]

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

The identities that'll help you are:

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

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

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

See how you get on.

 

[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.

 

[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,
<br /> \begin{array}{l}<br /> x + iy = \sqrt(x^2+y^2)\exp(i\arctan(y/x))<br /> \end{array}<br />
Take logarithms on both sides and you get equation 2.

 

[pnaj]

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

Does anyone know another way?

 

[Zurtex]

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.

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.