This question might seem easier than I'm making it seem. But how many values are there for $(1+i)^{2/3}$? Do I let $z=(1+i)^{2/3}$ so that $z^3=2i$? I'm asked to write each in polar coordinates and in rectangular form and also sketch their location. Any help will be appreciated thank you.
Answer
Here is how you advance
$$ z^{3} =(1+i)^{2}= 2i = 2e^{{i\pi/2}} = 2 e^{{i\pi/2}} e^{i2k\pi} =2e^{i\pi/2+i2k\pi} $$
$$ \implies z = 2^{1/3}e^{\frac{i\pi/2+i2k\pi}{3}},\quad k = 0,1,2. $$
No comments:
Post a Comment