Determine the trigonometric form of z=-2i.
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We have to determine the trigonometric form of z = -2i
The absolute value of z = -2i
sqrt [ 0^2 + (-2)^2]
=> sqrt 4
=> 2
and the the argument is -90 as the real component is zero.
Therefore the required result is 2 , -90 degrees
z = -2i. To find the trigonometric form
We know that if z = x+yi is the Cartesian form, then r*cost + i*sint is the trigonometric form of x+yi. Here r = (x^2+y^2)^(1/2) , rcost = x, r*sin t = y and t = arc tan (y/x).
Therefore z = 0+(-2*i)
r = {0^2+(-2)^2}^(/2) = 2.
x = 0 = 2cost and y= -2i = 2i*sint. So t = arc tan (-2/0) = -pi/2.
So -2i = 2cos(-pi)/2+ i*sin (-pi/2).
So 2cos(-pi)/2+ i*sin (-pi/2) is the trigonometric form of -2i.
Since the original form of the complex number is rectangular form, we'll transform it into the polar form.
z = a + bi
z = -2i
Re(z) = 0 and Im(z) = -2
The polar form:
z = |z|(cos t + i sin t)
|z| = sqrt[Re(z)^2 + Im(z)^2]
|z| = sqrt (0 + 4)
|z| = 2
tan t = Im (z)/Re(z)
Tan t = -2/0
t = -pi/2 or t = 3pi/2
The polar (trigonometric) form of the complex number is:
z = 2(cos 3pi/2 + i*sin 3pi/2)
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