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FP2 Urgent help

Solve the equation z^(3/4) = root6 + (root2)i, giving your answers in the form re^itheta, where r>0 and theta is more than -pi and less than or equal to pi.

OP

Can you guys reply with a fully worked solution?
I dont understand when to add the 2kpi...

Kvothe the Arcane

Original post by Pokepal03

Solve the equation z^(3/4) = root6 + (root2)i, giving your answers in the form re^itheta, where r>0 and theta is more than -pi and less than or equal to pi.

Demoirves theorum

[video]https://youtu.be/N0Y8ia57C24[/video]

OP

Original post by Kvothe the arcane

Demoirves theorum

[video]https://youtu.be/N0Y8ia57C24[/video]

Thank you! I understand everything, but when there isnt an integer above z, i dont get how to solve it... can you please elaborate?

Kvothe the Arcane

Original post by Pokepal03

Thank you! I understand everything, but when there isnt an integer above z, i dont get how to solve it... can you please elaborate?

Do it as normal and you'll have z^{1/4}=this for k=1, 2 etc...
Then you can say z=this for k=1, 2 etc.. via DeMoirves

Original post by Pokepal03

Can you guys reply with a fully worked solution?
I dont understand when to add the 2kpi...

Full solutions aren't encouraged on this forum - it doesn't help learning.

As for the

2k\pi

part, you know that cosine and sine are periodic functions, hence we can say that

\displaystyle \cos\theta \equiv \cos(2k\pi + \theta)

Write the complex number in exponential form (or polar form if you'd prefer - although arguably exponential form makes the algebra more intuitive -

\sqrt{6}+\sqrt{2}i \equiv \sqrt{8}e^{i\arctan{1/ \sqrt{3}}} \equiv \sqrt{8}e^{\pi i/6}

Does this perhaps make it easier to solve?

So

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

.

(edited 8 years ago)

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