# Find the polar form of the complex number 7 – 5i.

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### 3 Answers

To convert a complex number a + bi into the polar form we need to find r which is equal to sqrt (a^2 + b^2) and the angle theta given as arc tan (b/a)

For the complex number 7 - 5i we have a = 7 and b = -5

Therefore r = sqrt (7^2 + -5^2) = sqrt (49 + 25) = sqrt 74 = 8.6 (approximately)

theta = arc tan (-5/7) = -35.53 degree. We see that 7 – 5i lies in the fourth quadrant.

**The required polar form of 7 – 5i is 8.6 (cos -35.53 + i*sin -35.53).**

The polar form of x+yi = r(cost+isint), where r = sqrt(x^2+y^2).

cost = x/sqrt(x^2+y^2) , sint = y/(x^2+y^2).

r is called the modulus of x+yi and t is called the aruement or ampitude of x+yi.

Therefore the given complex number, x+yi = 7 - 5i

r = sqrt {(7^2+(-5)^2} = sqrt(25+49) = sqrt74.

Therefore cos t = 7/sqrt74 and sin t =-5/sqrt74.

Therefore t = arc tan (-5/7) = -0.6204 rad appr.

Therefore 7-5i = sqrt74{ cost+isint) , where t = -0.6204 rad.

7-5i = sqrt74({cos (arc tan (-5/7) +i sin arctan(-5/7)} in polar form.

The polar form of a complex number cold be found from the rectangular form;

z = 7 – 5i

The polar form is:

z = |z|(cos a + i*sin a)

|z| = sqrt [Re(z)^2 + Im(z)^2]

tan a = Re(z)/Im(z)

We'll identify Re(z) and Im(z) from the rectamgular form:

Re(z) = 7 and Im(z) = -5

|z| = sqrt(7^2 + 5^2)

|z| = sqrt(49+25)

|z| = sqrt 74

tan a = -7/5 => a = -arctan 7/5

The polar form of z is:

z = sqrt 74[cos (-arctan 7/5) + i*sin (-arctan 7/5)]

**z = sqrt 74[cos (arctan 7/5) - i*sin (arctan 7/5)]**