Question

write the cube root of i in x + yi form

Answer 1

t's incredibly ugly, since the form for (a+bi)3 is:

a3 + 3a2bi - 3ab2 - b3i

So if you're looking for the cube root of x + iy, you get

x = a3 - 3ab2
y = 3a2b - b3

And you want to solve for a and b. Maybe there's a neat trick, but I don't want to be the one to find it :P

Answer 2

\[\sqrt[3]{i}=x+ iy\]
now do
\[i=(x+iy)^3\]
solve for real and imaginary no.

Answer 3

The trick is to do it using polar coordinates. Whenever you cube a complex number, you triple its angle, and you cube its modulus. Since the modulus of \(i\) is 1, the three cube roots of \(i\) will also have modulus 1, just at different angles. What is the polar angle (also known as argument) of \(i\), and what three angles in the interval \([0,2\pi)\) will give you that angle when tripled?

Answer 4

This is what I mean by angle. Note that \(\tan \theta = \frac{b}{a}\). This should nudge you in the right direction.