Express the following in the form

*a *+ *ib*. Give all values and make a polar plot of

the points or the vector vectors that represent your result.

(a) (9*i*)^(1/2)

is this the answer? =3*(cos pi/2 + i sin pi/2)

=3i

so how do we represent this on polar plot?

(b)(-64*i*)^(1/4)=(64^0.25)*(cos theta/4 + i sin theta/4)
here im getting theta= -pi and +pi so which one do we take?

You need to remember that there are exactly two square roots, and they are evenly spaced around the unit circle. In other words, they differ by an angle of \(\displaystyle \pi\).

Convert \(\displaystyle 9i\) to polars:

\(\displaystyle |9i| = 9\)

\(\displaystyle \arg{(9i)} = \frac{\pi}{2}\).

So \(\displaystyle 9i = 9\,\textrm{cis}\,\frac{\pi}{2}\).

Therefore \(\displaystyle (9i)^{\frac{1}{2}} = \left(9\,\textrm{cis}\,\frac{\pi}{2}\right)^{\frac{1}{2}}\)

\(\displaystyle = 9^{\frac{1}{2}}\,\textrm{cis}\,\left(\frac{1}{2}\cdot \frac{\pi}{2}\right)\) By DeMoivre's Theorem

\(\displaystyle = 3\,\textrm{cis}\,\frac{\pi}{4}\).

And remembering that the other square root only differs by an angle of \(\displaystyle \pi\), that means the other square root is

\(\displaystyle 3 \,\textrm{cis}\,\frac{5\pi}{4}\).

Converting back to Cartesians:

\(\displaystyle 3\,\textrm{cis}\,\frac{\pi}{4} = 3\cos{\frac{\pi}{4}} + 3i\sin{\frac{\pi}{4}}\)

\(\displaystyle = \frac{3\sqrt{2}}{2} + i\,\frac{3\sqrt{2}}{2}\).

\(\displaystyle 3\,\textrm{cis}\,\frac{5\pi}{4} = 3\cos{\frac{5\pi}{4}} + 3i\sin{\frac{5\pi}{4}}\)

\(\displaystyle = -\frac{3\sqrt{2}}{2} - i\,\frac{3\sqrt{2}}{2}\).