Given:

Z= sqrt. of 2(cos(225degrees) +i sin(225degrees))^7

This the formula I'm using

Z^n= r^n(cos n deita + i sin n deita)

I tried to follow the formula but i keep getting it wrong.

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- Jul 30th 2008, 08:39 AMCyberman86De Moivre's Theorem finding powers
**Given:**

Z= sqrt. of 2(cos(225degrees) +i sin(225degrees))^7

This the formula I'm using

Z^n= r^n(cos n deita + i sin n deita)

I tried to follow the formula but i keep getting it wrong. - Jul 30th 2008, 09:25 AMTheEmptySet
The formula is correct. Note that

$\displaystyle 225\cdot 7 =1575=4\cdot 360+135$

so this tell us that $\displaystyle 1575^\circ$ and $\displaystyle 135^\circ $ are coterminal angles

$\displaystyle z^7=2^7\left( \cos(135^\circ)+i\sin(135^\circ)\right)=2^7\left( -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)=2^6(-\sqrt{2}+i\sqrt{2})$ - Jul 30th 2008, 09:27 AMPlato
Are you missing this?

$\displaystyle 7\left( {225^ \circ } \right) = 1575^ \circ \approx 135^ \circ $