write the following numbers in the polar form

Question

Write the following in the polar form re^iθ, -pie<0≤pie

a)πi (piei)

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=

 

Answer 1

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 )