Solution to z^(2/3) when z=1+i

[smutangama]

Homework Statement

Find all solutions to the equation

Homework Equations

z^(2/3)

z=1+i

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

 

[janhaa]

look like you are correct...

 

[Mark44]

Homework Statement

Find all solutions to the equation

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 z^2/3 when z = 1 + i?

Homework Equations

z^(2/3)

z=1+i

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

 

[HallsofIvy]

Homework Statement

Find all solutions to the equation

Homework Equations

z^(2/3)

z=1+i

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.

The Attempt at a Solution

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

Yes, that is the correct polar form.

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

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.