# Homework Help: Solution to z^(2/3) when z=1+i

1. Aug 27, 2013

### smutangama

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

Find all solutions to the equation

2. Relevant equations

z^(2/3)

z=1+i

3. The attempt at a solution

I put it in polar form and got √2e(∏/4)i

Then took the power of that while adding 2∏ni and got

(√2e(∏/4) + 2∏ni)^(2/3)

so the argument here is ∏/6 + (4∏ni)/3

The argument according to the mark sheet is ∏/6 + (2∏n/3)i

where did i go wrong?

thanks

2. Aug 27, 2013

### janhaa

look like you are correct...

3. Aug 27, 2013

### Staff: Mentor

What equation? The only equation you show is z = 1 + i. In case you're not clear on this, an equation has an equals sign (=) separating two expressions.

Is the problem to evaluate z2/3 when z = 1 + i?

4. Aug 27, 2013

### HallsofIvy

Either "find z^(2/3) where z= 1+ I" or "solve the equation "z^(3/2)= 1+ I". That's what Mark44 was complaining about.

Yes, that is the correct polar form.

You are correct that the "general" form of 1+ ix is $\sqrt{2}e^{ix(\pi/4+ 2n\pi)}$ so that the 2/3 power has argument $(2/3)(\pi/4+ 2n\pi)= \pi/6+ 4n\pi/3$.

If the problem is as stated, the "mark sheet" is incorrect.