Finding the nth root of a complex number?

[21joanna12]

Homework Statement

Find the solutions to z^{\frac{3}{4}}=\sqrt{6}+\sqrt{2}i

Homework Equations

de Moivre's theorem

The Attempt at a Solution

z^{\frac{3}{4}}=2\sqrt{2}e^{\frac{\pi i}{6}}=2\sqrt{2}e^{\frac{\pi i}{6}+2k\pi}=2\sqrt{2}e^{\frac{\pi +12k\pi}{6}i}

z=4e^{\frac{4}{3}{\frac{\pi +12k\pi}{6}i}=4e^{\frac{2\pi +24k\pi}{9}i}

For answers with a principal argument,

-\pi<\frac{2\pi +24k\pi}{9}\leq\pi
-\frac{11}{24}<k\leq\frac{7}{24}

So the only integer value of k satisfying this is zero. So I get z=4e^{\frac{2\pi}{9}i} as my only solution. However 4e^{\frac{8\pi}{9}i} and 4e^{\frac{-4\pi}{9}i} are also solutions...

Can anyone see where I have lost my solutions?

Thank you in advance :)

 

[Mentallic]

Just fixing up your post since one of your latex expressions had a syntax error.

Homework Statement

Find the solutions to z^{\frac{3}{4}}=\sqrt{6}+\sqrt{2}i

Homework Equations

de Moivre's theorem

The Attempt at a Solution

z^{\frac{3}{4}}=2\sqrt{2}e^{\frac{\pi i}{6}}=2\sqrt{2}e^{\frac{\pi i}{6}+2k\pi i}=2\sqrt{2}e^{\frac{\pi +12k\pi}{6}i}

z=4e^{\frac{4}{3}\frac{\pi +12k\pi}{6}i}=4e^{\frac{2\pi +24k\pi}{9}i}

For answers with a principal argument,

-\pi<\frac{2\pi +24k\pi}{9}\leq\pi
-\frac{11}{24}<k\leq\frac{7}{24}

So the only integer value of k satisfying this is zero. So I get z=4e^{\frac{2\pi}{9}i} as my only solution. However 4e^{\frac{8\pi}{9}i} and 4e^{\frac{-4\pi}{9}i} are also solutions...

Can anyone see where I have lost my solutions?

Thank you in advance :)

You're missing solutions because you introduced 2k\pi i too soon.

z^{3/4}=\sqrt{8}e^{\frac{\pi i}{6}}

z^3=64e^{\frac{2\pi i}{3}}=64e^{\frac{2\pi i}{3}+2k\pi i}

and now continue from there.

The reason it doesn't work your way is because you begin with 2k\pi i but then you raise each side to the 4th power, hence multiplying 2k\pi i by 4, giving you 8k\pi i which is obviously missing possible solutions.

 

[Ray Vickson]

Homework Statement

Find the solutions to z^{\frac{3}{4}}=\sqrt{6}+\sqrt{2}i

Homework Equations

de Moivre's theorem

The Attempt at a Solution

z^{\frac{3}{4}}=2\sqrt{2}e^{\frac{\pi i}{6}}=2\sqrt{2}e^{\frac{\pi i}{6}+2k\pi}=2\sqrt{2}e^{\frac{\pi +12k\pi}{6}i}

z=4e^{\frac{4}{3}{\frac{\pi +12k\pi}{6}i}=4e^{\frac{2\pi +24k\pi}{9}i}

For answers with a principal argument,

-\pi<\frac{2\pi +24k\pi}{9}\leq\pi
-\frac{11}{24}<k\leq\frac{7}{24}

So the only integer value of k satisfying this is zero. So I get z=4e^{\frac{2\pi}{9}i} as my only solution. However 4e^{\frac{8\pi}{9}i} and 4e^{\frac{-4\pi}{9}i} are also solutions...

Can anyone see where I have lost my solutions?

Thank you in advance :)

To find all the roots of $z^p = W$, take anyone root, $r$ and then multiply it by the $p$th roots of unity. Thus, if $\omega \neq 1$ is a root of $\omega^p = 1$, then so are $\omega^{-1}, \omega^{\pm 2}, \omega^{\pm 3}, \ldots$. That gives the other roots of $W$ as $r \omega^{\pm 1}, r \omega^{\pm 2}, \ldots$.