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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