Question

Find each principal root. Express the result in the form a+bi with a and b rounded to the nearest hundredth.
(32(cos2pi/3 +isin2pi/3))^1/5

Answer 1

Use De Moivre's Theorem.
\[r(\cos(\theta)+isin(\theta))^n = r^n(\cos(n \theta)+isin(n \theta))\]
after that expand and separate real and imaginary parts.

Answer 2

Ok

Answer 3

can I see the steps?

Answer 4

\[(32cis(2\pi/3))^\frac{ 1 }{ 5 } = 32^\frac{ 1 }{ 5 } cis(2\pi/15) = 2cis(2\pi/15)\]
Note cis is cos + isin for short. Then which part has the i in it? that is the b in a+bi and the other bit is a.

Answer 5

yeah,um how did you get the 15?

Answer 6

remember De Moivre's Theorem. It says to find the new argument, the power, n is multiplied by the argument when you raise any complex number in mod arg form to the power of n.

Answer 7

so \[\arg(z) = 2\pi/3; \arg(z ^{\frac{ 1 }{ 5 }}) = \frac{ 2\pi }{ 3 } \frac{ 1 }{ 5 } = \frac{ 2\pi }{ 15 }\]

Answer 8

oh ok

Answer 9

if you are stuck, recommend you go back and learn that theorem, De Moivre's Theorem. It is vital for complex numbers and it is a very easy formula actually.

Answer 10

ok thanks! @mww

Answer 11

I got 2+0.2i,I think that is wrong though.

Answer 12

nup you can't possibly get that cos(2pi/15) and sin(2pi/15) are irrational. Put in 2cos(2pi/15) in your calculator and 2sin(2pi/15) and that will be your real and imaginary parts

Answer 13

i still get the same answer sorry