complex number

Hello friends,
cant find the root here!
: Find the sixth roots of (12+5i)
Thanks,
Harshal

Quote:

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Originally Posted by harshal54321

Hello friends,
cant find the root here!
: Find the sixth roots of (12+5i)
Thanks,
Harshal

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convert the number to polar form:

12 + 5i = 13 e^(i [arctan(5/12) + 2 pi N]), N=0, +/-1. +/-2, ..

then the sixth roots are:

(12 + 5i)^(1/6) = (13)^(1/6) e^(i [arctan(5/12) + 2 pi N]/6), N=0, +/-1. +/- 2, ..

Which give distinct values for N= 0, 1, 2, 3, 4, 5.

RonL

Hello, Harshal!

I assume you're familiar with DeMoivre's Theorem . . .

Quote:

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Find the sixth roots of (12 + 5i)

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Convert to polar form: .12 + 5i .= .13(cosθ + i·sinθ), .where θ = arctan(5/12) .*

Then: .(12 + 5y)^(1/6) . = . 13^(1/6) (cos[θ + 2kπ)/6] + i·sin[(θ + 2kπ)/6])

. . Evaluate for k = 0, 1, 2, 3, 4, 5

*
Edit: corrected my typo.
. . . .Thanks for pointing it out, Captain.

Quote:

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Originally Posted by Soroban

Hello, Harshal!

I assume you're familiar with DeMoivre's Theorem . . .


Convert to polar form: .12 + 5i .= .13(cosθ + i·sinθ), .where θ = arctan(5/13)

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cos(theta)=12/13, sin(theta)=5/13, so tan(theta)=5/12

RonL