1. Use DeMoivre's theorem to find all the indicated roots. Find the three cube roots of 3 - 4i
2. One cube root of -1 is (1/2) - (1/2)*i*square root of 3. Cube this number in rectangular form and show that the result is -1.
The three cube roots of 3-4i are located on a circle with radius $\displaystyle \sqrt[3]{5}$, the real cube root of the absolute value of 3-4i, at three equal arcs. So each arc is measured at $\displaystyle \frac{{2\pi }}{3}$. So let $\displaystyle \Theta = \arg (3 - 4i) = \arctan \left( {\frac{{ - 4}}{3}} \right)$ then one root is $\displaystyle \sqrt[3]{5}\left( {\cos (\Theta ) + i\sin (\Theta )} \right)$
What are the other two?
well my teacher got 5 as the radius which i get and 306.87 degrees as the angle theta (might be mispelled). but then he goes from there to 5^(1/3)e^(306.87 degrees*i) sorry if this is cluttered. then he put that in polar form. but somewhere along the road he did the angle theta + 360/ 3 which got 222.3 degrees. from there i guess he got the other 2 roots. but im just a little confused in the methods used here. Sorry to ask but can you break it down into a step by step process, if possible? sorry i should be using latex for this stuff.
What is confusing you about this one? It's not particularly nice, but neither is it a conceptual challenge.
$\displaystyle \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right )^3$
$\displaystyle = \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right ) \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right ) \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right )$
$\displaystyle = \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right ) \left ( \left [ \frac{1}{2} \right ] ^2 - 2 \cdot \frac{1}{2} \cdot \frac{i}{2}\sqrt{3} + \left [ \frac{i}{2}\sqrt{3} \right ] ^2 \right )$
$\displaystyle = \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right ) \left ( \frac{1}{4} - \cdot \frac{i}{2}\sqrt{3} - \frac{3}{4} \right )$
$\displaystyle = \left ( \frac{1}{2} - \frac{i}{2}\sqrt{3} \right ) \left ( -\frac{1}{2} - \cdot \frac{i}{2}\sqrt{3} \right )$
Now you multiply in the other factor.
-Dan