1. ## Trignotmatric Problem

find all solution of Z^3/2=(1/2-i).

2. Originally Posted by winsome
find all solution of Z^3/2=(1/2-i).

Is the left hand side $\displaystyle z^{\frac{3}{2}}$ or $\displaystyle \frac{z^3}{2}$?

Is the right hand side $\displaystyle \frac{1}{2} - i$ or $\displaystyle \frac{1}{2-i}$?

3. Originally Posted by Prove It

Is the left hand side $\displaystyle z^{\frac{3}{2}}$ or $\displaystyle \frac{z^3}{2}$?

Is the right hand side $\displaystyle \frac{1}{2} - i$ or $\displaystyle \frac{1}{2-i}$?

I re-write this problem as below
Find all solution of Z^(3/2)=[(1/2) - i]

4. $\displaystyle z^{\frac{3}{2}} = \dfrac{1}{2}-i \Leftrightarrow z = \left(\dfrac{1}{2}-i\right)^{\frac{2}{3}}.$

5. Originally Posted by TheCoffeeMachine
$\displaystyle z^{\frac{3}{2}} = \dfrac{1}{2}-i \Leftrightarrow z = \left(\dfrac{1}{2}-i\right)^{\frac{3}{2}}.$
Actually it's

$\displaystyle \left(\frac{1}{2} - i\right)^{\frac{2}{3}}$.

And this is only one solution. There will be two others.

Convert to polars, use DeMoivre's Theorem to simplify, then recall that the three roots are evenly spaced around a circle to find the other two.

6. Originally Posted by Prove It
Actually it's

$\displaystyle \left(\frac{1}{2} - i\right)^{\frac{2}{3}}$.
Thanks. That's what I had thought I have written.