Question

Use De Moivre's Thereom to compute (sqrt(3) + i)^4.

Answer 1

So if \( z = \sqrt{3} + i \), what is \( z \) in polar form, \[ z = re^{i\theta} \] That is, what are \( r \) and \( \theta \).

Answer 2

Once you know that,
\[ z^4 = r^4 e^{4\theta i} \]

Answer 3

Is it (r cos ø + r sin (ø)i ) ?

Answer 4

in polar form?

Answer 5

Right, same thing it turns out,
\[ z = r( \cos \theta + i \sin \theta ) \]
What value of r and theta gives your value of z?

Answer 6

Then \( z^4 = r^4 ( \cos(4\theta) + i \sin(4\theta)) \).

Answer 7

Ok, i think that \[r = \sqrt{a ^{2} + b ^{2}} = 2\]

Answer 8

yes

Answer 9

whenever u are free @james
http://openstudy.com/#/updates/4ed58b7be4b0bcd98ca6ebe8

Answer 10

but i'm having trouble understanding how to find theta

Answer 11

Draw a diagram and think about the tan of theta. I'm going to go to this other problem for a bit.

Answer 12

ok :)

Answer 13

found theta?

Answer 14

i'm not sure, but i think its arctan of a/b, so arctan of \[1/\sqrt{3}\] ?

Answer 15

so i think pi/6?

Answer 16

Yes. And you should draw the diagram of z in the complex plane to convince yourself that's right.

You should also see where that formula comes from. See it geometrically in your diagram.

And then see it algebraically. If
z = a + bi
= r . cos(theta) + r. sin(theta) i

Then

b/a = sin(theta)/cos(theta) = tan(theta)
=> theta = arctan(b/a)

For the record, you have to be quite careful with this formula. For example if z = 1 + i, then tan(theta) = 1. And if w = -1 - i, then tan(theta) also equals 1. But the theta are quite different.

Answer 17

Anyway, for your z, tan(theta) = 1/sqrt(3). As theta lies in the first quadrant, it must be that theta = pi/6.

Answer 18

OK, Thanks! I think I can handle it from there. :)