# Thread: Complex number.. need an idea to start..

1. ## Complex number.. need an idea to start..

1. The problem statement, all variables and given/known data

Find the principal value of i^2i

2. Relevant equations

principal value = Ln r + iArg(z)

3. The attempt at a solution
dont know how to start..

2. Originally Posted by nameck
1. The problem statement, all variables and given/known data

Find the principal value of i^2i

2. Relevant equations

principal value = Ln r + iArg(z)

3. The attempt at a solution
dont know how to start..
Are you asking for the principle value of $\ln(i^{2i})$? If so it would be nice if you say so.

(start by writing $i$ in exponetial form: $i=r e^{i\theta}$, with $r=1$ and $\theta = \pi/2 + 2n\pi$)

CB

3. I think we are to find the principle value of $i^{2i}$.
$
i^{2i} = \exp \left( {2i\text{Log}(i)} \right) = \exp \left( {(2i)\left( {i\frac{\pi }{2}} \right)} \right) = ?$

4. Yes Plato u are right..

5. I dif that for you. Can you finish?

6. Originally Posted by Plato
I think we are to find the principle value of $i^{2i}$.
$
i^{2i} = \exp \left( {2i\text{Log}(i)} \right) = \exp \left( {(2i)\left( {i\frac{\pi }{2}} \right)} \right) = ?$
how did u get pi/2?

7. Originally Posted by nameck
how did u get pi/2?
The principle value of $z^w=\exp\left(w\cdot\text{Log}(z)\right)$.
Now $\text{Arg}(i)=\frac{\pi}{2}$ and $|i|=1$ so $\ln(1)=0$

8. Originally Posted by Plato
I think we are to find the principle value of $i^{2i}$.
$
i^{2i} = \exp \left( {2i\text{Log}(i)} \right) = \exp \left( {(2i)\left( {i\frac{\pi }{2}} \right)} \right) = ?$
when did u insert Arg(z)?