Determine the trigonometric form of z=-2i.

3 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)

We’ve answered 318,944 questions. We can answer yours, too.

Ask a question