Question

Which expression is a sixth root of -1+isquare root3? Please HELP!!

Answer 1

@zepdrix

Answer 2

oo fun problem =o

Answer 3

\[\Large\rm z=-1+ \mathcal i\sqrt3\]Recall how we find the radial length,\[\Large\rm r=\sqrt{(-1)^2+(\sqrt3)^2}\]
And to find the angle of a complex number: \(\Large\rm x+\mathcal i y\)
we use\[\Large\rm \tan \theta=\frac{y}{x}\]

So for our problem:\[\Large\rm \tan \theta=\frac{\sqrt3}{-1},\qquad\to\qquad \tan \theta=-\sqrt3\]

Answer 4

This value corresponds to two special angles in one rotation around the unit circle ( only one of those angles is in the correct quadrant though ).
Look back at your z to figure out which quadrant you're in.

Answer 5

If you're still unsure of how to find the angle, you can use your calculator,\[\Large\rm \tan \theta=-\sqrt3 \qquad\to\qquad \tan^{-1}(-\sqrt3)=\theta\]

Answer 6

Make sure you're in degree mode.

Answer 7

The inverse tangent function can only spit out angles in the first and fourth quadrants.
So for example, in this problem: Our value is negative, tangent is negative in quadrants two and four.

So your calculator will give you the angle in quadrant four.
You might have to ADD 180 degrees to find the angle in quadrant two if that's the one you're looking for.

Answer 8

Oh oh oh, I'm such a doofus...
we need to take the 6th root of that value.

So after we get our polar form,\[\Large\rm z=r\left(\cos \theta+\mathcal i \sin \theta\right)\]To find one of the six roots we'll simply take the 6th root of each side and then apply De'Moivre Theorem.
\[\Large\rm z^{1/6}=\left[r\left(\cos \theta+\mathcal i \sin \theta\right)\right]^{1/6}\]\[\Large\rm z^{1/6}=r^{1/6}\left(\cos \theta+\mathcal i \sin \theta\right)^{1/6}\]De'Moivre's Theorem tells us that we can bring the power down and multiply it by our angle,\[\Large\rm z^{1/6}=r^{1/6}\left(\cos \left[\frac{1}{6}\theta\right]+\mathcal i \sin \left[\frac{1}{6}\theta\right]\right)\]