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November 22, 2014

November 22, 2014

Posted by **Abbey(Please help)** on Tuesday, May 4, 2010 at 4:58pm.

(2+2i)^6

a=2 b=2 n=6

r=sqrt 2^2 + 2^2 = sqrt8

Q=7pi/4

(sqrt8)^6 = 512

512(cos 6/1 x 7pi/4) + i sin 6 x 7pi/4

512 (cos 21pi/2) + i sin (21pi/2)

512(cos pi/2) + i (sin pi/2)

512(2+2i)

1024 + 1024i

Is this correct?

- Math -
**gorge**, Tuesday, May 4, 2010 at 5:10pmi=253

- Math -
**drwls**, Tuesday, May 4, 2010 at 5:34pmRewrite (2+2i)^6 as

[2^3/2*(cos pi/4 + i sin pi/4)]^6

= 2^9*[cos 6pi/4 * i sin 6pi/4]

= 512* i sinpi/2

= -512 i

I see several error in your derivation.

- Math -
**Reiny**, Tuesday, May 4, 2010 at 5:34pmWhere did you get the angle of 7π/4 ?

I get an angle of π/4

check:

√8(cos π/4 + i sin π/4_

= √8(1/√2 + i (1/√2)

= 2 + 2i

so by De Moivre's Theorem

(2+2i)^6

= √8^6(cos (6π/4) + i sin (6π/4))

= 512( + i(-1))

= - 512i

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