Math CentralQuandaries & Queries


Subject: Exponential form
Name: Austin
Who are you: Student

When dealing with imaginary numbers in engineering, I am having trouble getting things into the exponential form. The equation is -1+i now I do know that
re^(theta)i = r*cos(theta) + r*i*sin(theta). Just not quite understanding the order of operations. Thanks

Hi Austin,

To express -1 + i in the form r eitheta = r (cos(theta) + i sin(theta)) I think of the geometry. In the complex plane plot the point -1 + i.

complex plane

The modulus r of p = -i + i is the distance from O to P. Since PQO is a right triangle Pythagoras theorem tells you that r = √2. The argument of P is the angle, theta measured counterclockwise from the positive real axis to the line segment OP. Here theta = 3/4 pi. Cos( 3/4 pi) = -1/√2 and sin(3/4 pi) = 1/√2. Thus

You don't need to use the diagram. The modulus of x + iy is √(x2 + y2) which for -i + i is

r = √((-1)2 + 12) = √2

Also x = r cos (theta) and y = r sin(theta) so -1 = √2 cos (theta) and 1 = √2 sin (theta). This gives theta = 3/4 pi.


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