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# FP2 Urgent help watch

1. 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.
2. Can you guys reply with a fully worked solution?
I dont understand when to add the 2kpi...
3. (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

https://youtu.be/N0Y8ia57C24
4. (Original post by Kvothe the arcane)
Demoirves theorum

https://youtu.be/N0Y8ia57C24
Thank you! I understand everything, but when there isnt an integer above z, i dont get how to solve it... can you please elaborate?
5. (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
6. (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 part, you know that cosine and sine are periodic functions, hence we can say that
7. Write the complex number in exponential form (or polar form if you'd prefer - although arguably exponential form makes the algebra more intuitive -
Does this perhaps make it easier to solve?

So .

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