Dalia S.

asked • 02/22/14# write the following in the polar form

^{eiθ}, -pie<0≤pie

a)πi

r=

θ=

b)-2√2- 5i

r=

θ=

c) (1-i)(-√(3)+i))

r=

θ=

d)(√(2)-5i))2

r=

θ=

e)(-5+√(2i))/(5+3i)

r=

θ=

f) -√7(1+i)/ √(3) +i

r=

θ=

## 2 Answers By Expert Tutors

Steve S. answered • 02/22/14

Tutoring in Precalculus, Trig, and Differential Calculus

http://www.wyzant.com/resources/answers/27705/

Here’s my answer, again:

[I would like other tutors to critique it for errors, though.]

Write the following in the polar form re^(iθ), -pi < 0 ≤ pi

a) pi i = re^(iθ), where:

r = pi

θ = pi/2

b) -2√2 - 5i = re^(iθ), where:

r = √(8+25) = √(33)

θ = -pi + arctan((5√(2))/4)

c) (1-i)(-√(3)+i)) = (√(2) e^(-i pi/4)) (2 e^(i 5pi/6))

= 2√(2) e^(i 7pi/12)

r = 2√(2)

θ = 7pi/12

d)(√(2)-5i))^2 = ( √(27) e^(-i arctan( (5√(2))/2) ) ) )^2

= 27 e^(-2i arctan( (5√(2))/2) ) )

r = 27

θ = -2 arctan( (5√(2))/2) )

e)(-5+√(2i))/(5+3i)

2i = 2e^(i pi/2)

√(2i) = √(2) e^(i pi/4) = 1 + i

-5+√(2i) = -4 + i = √(17) e^(i (pi + arctan(-1/4)))

5+3i = √(34) e^(i arctan(3/5))

(-5+√(2i))/(5+3i) = ( √(17) e^(i (pi + arctan(-1/4))) )/( √(34) e^(i arctan(3/5)) )

= (√(2))/2 e^(i ( (pi + arctan(-1/4))) - (arctan(3/5)) ) )

r = (√(2))/2

θ = pi + arctan(-1/4) - arctan(3/5) = pi + arctan(7/23)

f) -√7(1+i)/(√(3)+i)

- √7(1+i) = - √(7) √(2) e^(i pi/4)

√(3)+i = 2 e^(i arctan( (√(3))/3 ) )

-√7(1+i)/(√(3)+i) = - √(14) e^(i pi/4) / ( 2 e^(i arctan( (√(3))/3 ) ) )

r = (√(14))/2

θ = pi/4 - arctan( (√(3))/3 )

Dalia S.

02/22/14

Nathan C. answered • 02/22/14

Math and Science (K through College)

r = sqrt((-1)^2 + (+1)^2) = sqrt(1 + 1) = sqrt(2) -> r = sqrt(2)

θ = arctan((+1)/(-1)) = π*3/4

r = sqrt((-1)^2 + (-1)^2) = sqrt(1 + 1) = sqrt(2) -> r = sqrt(2)

θ = arctan((-1)/(-1)) = -π*3/4

r = sqrt((+1)^2 + (-1)^2) = sqrt(1 + 1) = sqrt(2) -> r = sqrt(2)

θ = arctan((-1)/(+1)) = -π/4

Steve S.

02/22/14

Steve S.

02/22/14

Nathan C.

02/22/14

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Nathan C.

02/22/14