# Complex number

• Oct 27th 2012, 09:00 AM
mathsforumhelp
Complex number
What is 3^(3+ i(8pi/ln3))

Work done so far

3^3 * 3^( i8pi/ln(3) ) = 27 * (3^ i8pi - 3^ln(3) ) = 27 *( 3^25.1i -3.34)

3^(3+ i25.1) - 90.26

I am not sure how to convert i to a number.........
• Oct 27th 2012, 09:30 AM
Plato
Re: Complex number
Quote:

Originally Posted by mathsforumhelp
What is 3^(3+ i(8pi/ln3))

I don't know how advanced you are.
In complex numbers $\displaystyle z^w=\exp(w\cdot \log(z))$.
Depending upon your level, you may want to use only the principal log.
If so, we get $\displaystyle \exp \left( {\ln (3)\left( {3 + \frac{{8\pi i}}{{\ln (3)}}} \right)} \right) = \exp \left( {\ln (27) + 8\pi i} \right) = 27$
• Oct 27th 2012, 12:03 PM
Soroban
Re: Complex number
Hello, mathsforumhelp!

Quote:

$\displaystyle \text{Evaluate: }\:X \;=\;3^{3+ \frac{8\pi}{\ln(3)}i}$

We have: .$\displaystyle X \;=\;3^3\cdot3^{\frac{8\pi}{\ln(3)}i} \;=\;27\cdot3^{\frac{8\pi}{\ln(3)}i}$ .[1]

Let $\displaystyle y \:=\:3^{\frac{8\pi}{\ln(3)}i}$

Take logs: .$\displaystyle \ln(y) \:=\:\ln\left(3^{\frac{8\pi}{\ln(3)}i}\right) \:=\:\frac{8\pi}{\ln(3)}i\cdot\ln(3) \quad\Rightarrow\quad \ln(y) \:=\:\8\pi i$

Hence: .$\displaystyle y \:=\:e^{8\pi i} \:=\:(e^{i\pi})^8 \:=\:(\text{-}1)^8 \:=\:1$

Substitute into [1]: .$\displaystyle X \:=\:27\cdot1 \:=\:27$

• Oct 27th 2012, 08:51 PM
mathsforumhelp
Re: Complex number
Thanks guys.....