- #1

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z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

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- Thread starter kasse
- Start date

- #1

- 366

- 0

z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

- #2

malawi_glenn

Science Advisor

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z^4= 1/2 + i sqrt(3)/2

I start by transforming into polar form:

z^4 = e^(i*Pi/3)

But then I'm blank.

Have you tried De Moivre's formula ?

- #3

CompuChip

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[tex] \sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2} [/tex]

- #4

malawi_glenn

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[tex] \sqrt{r e^{i \phi}} = \sqrt{r} e^{i \phi / 2} [/tex]

yes, and from De Moivre's formula we get the general:

[tex] z^{1/n} = r^{1/n}*exp(i \phi / n) [/tex]

- #5

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yes, and from De Moivre's formula we get the general:

[tex] z^{1/n} = r^{1/n}*exp(i \phi / n) [/tex]

So e^((i Pi)/12) is a solution. How about the other three?

- #6

malawi_glenn

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i.e arg((1/2) + i (rot3 / 2)) = pi / 3 + 2 pi * n, where n is ... -3,-2,-1,0,1,2,3...

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