Find the argument of the complex number (2+2i)^11/(2-2i)^9

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,995 questions. We can answer yours, too.

Ask a question