use Demoivre's Theorem to find the indicated power of the complex number. Express the result in standard form.
(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?
3 answers
i=253
Rewrite (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.
[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.
Where 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
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
0 answers