De Moivre's Theorem finding powers

**Cyberman86** · July 30th 2008, 08:39 AM

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.

**TheEmptySet** · July 30th 2008, 09:25 AM

Originally Posted by Cyberman86

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.

The formula is correct. Note that

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

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

$z^7=2^7\left( \cos(135^\circ)+i\sin(135^\circ)\right)=2^7\left( -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)=2^6(-\sqrt{2}+i\sqrt{2})$

**Plato** · July 30th 2008, 09:27 AM

Are you missing this?
$7\left( {225^ \circ } \right) = 1575^ \circ \approx 135^ \circ$

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