- #1

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I tried using the formula x^ni = cos (ln (x)^n) + i sin (ln (x)^n)

but i cannot solve it. i used matlab to get this answer 0.0432139182637723 + 0i

but i dunno how to solve it with steps.. can i get some assistance please.

thank you.

- Thread starter aks_sky
- Start date

- #1

- 55

- 0

I tried using the formula x^ni = cos (ln (x)^n) + i sin (ln (x)^n)

but i cannot solve it. i used matlab to get this answer 0.0432139182637723 + 0i

but i dunno how to solve it with steps.. can i get some assistance please.

thank you.

- #2

- 3,470

- 246

Do you know how to write -1 in polar form?

I tried using the formula x^ni = cos (ln (x)^n) + i sin (ln (x)^n)

but i cannot solve it. i used matlab to get this answer 0.0432139182637723 + 0i

but i dunno how to solve it with steps.. can i get some assistance please.

thank you.

- #3

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- 0

yup the polar form will just be cos (theta) + i sin (theta) and the modulus here is 1.. correct?

- #4

- 3,470

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Well, that's a particular complex number, but it's actually expressed in rectangular form x + iy, where x = cos(theta) and y = sin(theta).yup the polar form will just be cos (theta) + i sin (theta) and the modulus here is 1.. correct?

Do you know how to write -1 in terms of "e", i.e., do you know what a complex exponential is? It would help to know what background can be assumed for this exercise.

- #5

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which is.. i ln (-1)

then in terms of "e" i will get... e ^ i ln (-1)

which gives me cos (ln (-1)) + i sin (ln (-1))

but i cant go any further to get the answer

- #6

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What I was trying to get at is, have you been exposed to Euler's famous formula:

which is.. i ln (-1)

then in terms of "e" i will get... e ^ i ln (-1)

which gives me cos (ln (-1)) + i sin (ln (-1))

but i cant go any further to get the answer

[tex]e^{i\pi} = -1[/tex]

If so, then you can easily use this to get the answer you want.

- #7

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- #8

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Well, you're trying to find (-1)^i, right? So what is the natural thing do to both sides of Euler's formula?

- #9

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um not sure exactly.

- #10

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Oh, come on!um not sure exactly.

What operation do you do to -1 to obtain (-1)^i? (This isn't a trick question!) Just do that operation to both sides of Euler!

- #11

Matterwave

Science Advisor

Gold Member

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[tex](-1) = (e^{i\pi})[/tex]

[tex](-1)^i = ...[/tex]

Use basic algebra here.

- #12

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we take logs of both sides

- #13

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ohhh yup i get what you asking

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