Use DeMoivre's Theorem to write the complex number in trigonometric form. (sqrt2 + I sqrt2)^3
@ zepdrix
@zepdrix
So we need 2 pieces of information: The magnitude, \(\Large\rm r\), of our thing inside the brackets, and the angle it forms \(\Large\rm \theta\).
So when we have the form:\[\Large\rm a+b\mathcal i\](Which we do) Then the radius is given by: \[\Large\rm r=\sqrt{a^2+b^2}\]
so that would be 2
Mmm ok good.
r=2
Recall that tangent is \(\Large\rm \dfrac{opposite}{adjacent}\) What that means for us here is that the tangent of theta is the imaginary component (opposite side) over the real component (adjacent side). \[\Large\rm \tan \theta=\frac{b}{a}\]
Website crashed again? :( ugh great...
that equals 1
@zepdrix you still there
\[\Large\rm \tan \theta=1\]Yes, good. So how do we solve for theta? What special angles does this correspond to?
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