What is the argument of (1+i)^13/(1-i)^7 ?

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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)

The reason of putting the numerator and denominator in polar form is to use Moivre's rule.

We'll put the numerator in polar form:

z1 = 1+i

Re(z1)  =1

Im(z1) = 1

r1 = sqrt(1^2 + 1^2)

r1 = sqrt2

tan t = Im(z1)/Re(z1)

tan t = 1/1

t = arctan 1

t = pi/4

z1 = sqrt2(cos pi/4 + i*sin pi/4)

(1+i)^13 = [sqrt2(cos pi/4 + i*sin pi/4)]^13

We'll use Moivre's rule:

(1+i)^13 = 2^(13/2)(cos 13pi/4 + i*sin 13pi/4)

(1+i)^13 = 2^(13/2)(cos 5pi/4 + i*sin 5pi/4)

We'll put the denominator in polar form:

z2 = 1 - i

Re(z2)  =1

Im(z2) = -1

r2 = sqrt(1^2 + (-1)^2)

r2 = sqrt2

t2 = arctan -1

t2 = -pi/4

z2 = sqrt2(cos -pi/4 + i*sin -pi/4)

(1-i)^7 = [sqrt2(cos -pi/4 + i*sin -pi/4)]^7

We'll use Moivre's rule:

(1-i)^7 = 2^(7/2)*(cos -7pi/4 + i*sin -7pi/4)

(1-i)^7 = 2^(7/2)*(cos (2pi-7pi/4) + i*sin (2pi-7pi/4))

(1-i)^7 = 2^(7/2)*(cos pi/4 + i*sin pi/4)

Now, we'll calculate the ratio:

(1+i)^13/(1-i)^7 = 2^[(13-7)/2]*[cos (5pi-pi)/4 + i*sin (5pi-pi)/4]

(1+i)^13/(1-i)^7 = 2^3* (cos pi + i*sin pi)

Since cos pi = -1 and sin pi = 1, we'll get:

(1+i)^13/(1-i)^7 = -2^3

(1+i)^13/(1-i)^7 = -8

The argument of the ratio (1+i)^13/(1-i)^7  is pi.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find  the argument of (1+i)^13/(1-i)^7.

Let simplify Z = (1+i)^13/(1-i)^7 beore finding the argument of z.

z = (1+i)^13/(1-i)^7 = (1+i)^13* (1+i)^7/(1-i)^7*(1+i)^7.

z = (1+i)^20 /({1- i^2}^&

z = (1+i)^20/(2)^7.

z = (sqr2 )^20 *{ 1/sqr2 + i/sqrt2}^2

z = {2^10/2^7} { cospi/4 +isin p/4}^20.

z = 2^3 { cos 5pi +isin5pi} , by D' Moivre's theorem.

Z = 2^3 { cospi + i sin pi}

 Therefore the argument of z = pi.

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