De Moivre's theorem problem :(

**scott.b89** · March 4th 2009, 11:00 AM

The question states:

Use the polar form and De Moivre's Theorem to simplify:
(a) (1+i)^5
_______

1 - i

So I categorised each as z1 and z2, worked out the modulus and argument of each, ( |z1| = root 2, |z2| = root 2, arg z1 = pie/4, arg z2 = - pie/4)

Carried out de moivre's theorem and got as far as the step

4cis(5pie/4)
__________

cis(-pie/4)

It would be much appreciated if anyone could help me finish it off or correct me if I've gone wrong somewhere along the line. Cheers

**HallsofIvy** · March 4th 2009, 11:18 AM

Originally Posted by scott.b89

The question states:

Use the polar form and De Moivre's Theorem to simplify:
(a) (1+i)^5
_______

1 - i

So I categorised each as z1 and z2, worked out the modulus and argument of each, ( |z1| = root 2, |z2| = root 2, arg z1 = pie/4, arg z2 = - pie/4)

Carried out de moivre's theorem and got as far as the step

4cis(5pie/4)
__________

cis(-pie/4)

Think [tex]\frac{x}{y}[tex] as $xy^{-1}$ and subtract those two arguments.

It would be much appreciated if anyone could help me finish it off or correct me if I've gone wrong somewhere along the line. Cheers

**scott.b89** · March 4th 2009, 11:27 AM

Do you think you could talk me trough it please. It's just that what you said there doesn't make sense to me as it's not a method I'm familiar with, are you American by chance, that might explain it :P

**HallsofIvy** · March 4th 2009, 11:37 AM

Originally Posted by scott.b89

Do you think you could talk me trough it please. It's just that what you said there doesn't make sense to me as it's not a method I'm familiar with, are you American by chance, that might explain it :P

What method are you familiar with? If $x= r cis(\theta)$ and $y= R cis(\phi)$ , what is $xy$ ?

If $x= r cis(\theta)$ and $y= R cis(\phi)$ , what is $\frac{x}{y}$ ?

**scott.b89** · March 4th 2009, 11:58 AM

What method are you familiar with? If

and

, what is

?

If

and

, what is

?

I've still never seen it expressed in that way before :P but...

x = 4cis(5pie/4)

y = cis(-pie/4)

I think?

**Plato** · March 4th 2009, 01:08 PM

Because $\frac{1}{{1 - i}} = \frac{{1 + i}}{2}$ we get $\frac{{(1 + i)^5}}{{1 - i}} = \frac{{\left( {1 + i} \right)^6 }}{2}$ .

We get $\left( {1 + i} \right)^6 = \left( {\sqrt 2 } \right)^6 cis\left( {\frac{{6\pi }}{4}} \right)$ .

Now you can finish.

**HallsofIvy** · March 4th 2009, 01:09 PM

Originally Posted by scott.b89

What method are you familiar with? If

and

, what is

?

If

and

, what is

?

I've still never seen it expressed in that way before :P but...

x = 4cis(5pie/4)

y = cis(-pie/4)

I think?

Not quite what I asked but okay, assuming that x and y are those, please answer my question.

If you do not know how to multiply or divide numbers in that form, how did you get the expression for the power?

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