Find the argument of the complex number (2+2i)^11/(2-2i)^9
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The complex number (2+2i)^11/(2-2i)^9 has to be simplified first.
(2+2i)^11/(2-2i)^9
=> (4 + 4i^2 + 8i)^5*(2 + 2i) / (4 +4i^2 - 8i)^4 * (2 - 2i)
=> (8i)^5*(2 + 2i) / (- 8i)^4 * (2 - 2i)
=> (64i^2)^5*(2 + 2i) / (64i^2)^4 * (2 - 2i)
=> -64*(2 + 2i) / (2 - 2i)
=> -64*(2 + 2i)(2 + 2i) / (2 - 2i)(2 + 2i)
=> -64*(2 + 2i)(2 + 2i) / (4 + 4)
=> -8*(2 + 2i)(2 + 2i)
=> -8*(4 + 4i^2 + 8i)
=> -64i
The argument of -64i is -90
The argument of the complex number is -90 degrees
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We'll write the numerator and denominator in polar form:z = r(cos t + i*sin t), to apply Moivre's rule, for finding the argument
We'll put the numerator in polar form:
z1 = 2+2i
Re(z1) =2
Im(z1) = 2
r1 = sqrt(2^2 + 2^2)
r1 = sqrt8
tan t = Im(z1)/Re(z1)
tan t = 2/2
t = arctan 1
t = pi/4
z1 = sqrt8(cos pi/4 + i*sin pi/4)
(2 + 2i)^11 = [sqrt8(cos pi/4 + i*sin pi/4)]^11
We'll use Moivre's rule:
(2 + 2i)^11 = 8^(11/2)*(cos 11pi/4 + i*sin 11pi/4)
(2 + 2i)^11 = 8^(11/2)*(cos 3pi/4 + i*sin 3pi/4)
We'll put the denominator in polar form:
z2 = 2 -2i
Re(z2) =2
Im(z2) = -2
r2 = sqrt(2^2 + (-2)^2)
r2 = sqrt8
t2 = arctan -1
t2 = -pi/4
z2 = sqrt8(cos -pi/4 + i*sin -pi/4)
(2-2i)^9 = [sqrt8(cos -pi/4 + i*sin -pi/4)]^9
We'll use Moivre's rule:
(2-2i)^9 = 8^(9/2)*(cos -9pi/4 + i*sin -9pi/4)
(2-2i)^9 = 8^(9/2)*(cos (2pi-9pi/4) + i*sin (2pi-9pi/4))
(2-2i)^9 = 8^(9/2)*(cos -pi/4 + i*sin -pi/4)
Now, we'll calculate the ratio:
(2+2i)^11/(2-2i)^9 = 8^[(11-9)/2]*[cos (3pi+pi)/4 + i*sin (3pi+pi)/4]
(2+2i)^11/(2-2i)^9 = 8* (cos pi + i*sin pi)
Since cos pi = -1 and sin pi = 0, we'll get:
(2+2i)^11/(2-2i)^9 = -8
The argument of the complex number (2+2i)^11/(2-2i)^9 is pi.
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