# Thread: Complex Powers of Complex Numbers

1. ## Complex Powers of Complex Numbers

Hey,
I have the complex number (-(root 3)+i)^i. and using complex number mathematics I have put the complex number into exponential form: 2^i * e^(-5pi/6).
However, this is not in exponential form as e^(-5pi/6) is not an imaginary number. Therefore the 2^i is the imaginary number. How do I get 2^i into exponential form using natural logs?

Thanks

2. Originally Posted by littlea86
I have the complex number (-(root 3)+i)^i. and using complex number mathematics However, this is not in exponential form as e^(-5pi/6) is not an imaginary number.
Here is the rule for complex exponentiation: $z^w=\exp(w\log(z))$.
So $(-\sqrt{3}+i)^i=\exp(i\log(-\sqrt{3}+i)$.
$\log(-\sqrt{3}+i)=ln(2)+ i \left( {\frac{{ - 5\pi }}{6} + 2n\pi })$.

3. $2^i= e^{ln(2^i)}= e^{i ln(2)}$

$(-\sqrt{3}+ i)^i= re^{i\theta}$ with $r= e^{-5\pi/6}$, as you say, and $\theta= ln(2)$.

4. Thank you so much for all your help

5. Originally Posted by littlea86
Hey,
How do I find (-(root3)+i)^i, expressing the answer in trigonometric form using exact values?

I have aready got the equation into the form 2cis5pi/6 but now i have to put it to the power of i (2cis5pi/6)^i, what do i have to do now?

Thanks
$\displaystyle z^{\alpha}=e^{\alpha \text{Ln} z}=e^{i \text{Ln} (-\sqrt{3}+i)}$

$\text{Ln}=\ln|z|+i\text{Arg}(z)$

$e^{-\text{Arg}(z)}\left(\cos(\ln(z))+i\sin(\ln(z))\rig ht)$

$\displystyle |z|=\sqrt{\left(\sqrt{3}\right)^2+1^2}=2, \ \ \text{Arg}(z)=\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{5\pi}{6}$