Thread: De Moivre's theorem problem :(

1. De Moivre's theorem problem :(

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

2. 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 $\displaystyle xy^{-1}$ 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

3. 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

4. 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 $\displaystyle x= r cis(\theta)$ and $\displaystyle y= R cis(\phi)$, what is $\displaystyle xy$?

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

5. 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?

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

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

Now you can finish.

7. 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?