Polarrepresentation of 1/(1/2 - 1/2i)?

[Humboldt]

Hello,
What's the Polarrepresentation of 1/(1/2 - 1/2i)?

My result was: √2e^(i(π/4)) , but the solution on my worksheet says 2e^(i(π/4)). Can someone explain me if I'm wrong or if they forget the root on the solution.

I'm sorry if I posted in the wrong category.
Thx in advance.

 

[MathLover1]

you are not wrong

1/(1/2 - 1/2i)

1(1/2 +1/2i)/((1/2 - 1/2i)(1/2 + 1/2i))

(1/2( 1+i))/((1/2)^2 - (1/2i)^2)

(1/2( 1+i))/(1/4 - 1/4*(i)^2)

(1/2( 1+i))/(1/4 - 1/4*(-1))

(1/2( 1+i))/(1/4 + 1/4)

(1/2( 1+i))/(2/4)

(1/2( 1+i))/(1/2)

1+i =>a+bi

in polar form:

r=sqrt(a^2+b^2)

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

r*cos(θ)=a =>sqrt(2)*cos(θ)=1

=>cos(θ)=1/sqrt(2)

θ=cos^-1(1/sqrt(2))

θ=π/4 (result in radians)

r*sin(θ)=b =>sqrt(2)*sin(θ)=1

=>sin(θ)=1/sqrt(2)

θ=sin^-1(1/sqrt(2))
θ=π/4

sqrt(2) (cos(π/4)+i* sin(π/4)) or sqrt(2) e^((i π)/4)

 

[nycmathguy]

you are not wrong

1/(1/2 - 1/2i)

1(1/2 +1/2i)/((1/2 - 1/2i)(1/2 + 1/2i))

(1/2( 1+i))/((1/2)^2 - (1/2i)^2)

(1/2( 1+i))/(1/4 - 1/4*(i)^2)

(1/2( 1+i))/(1/4 - 1/4*(-1))

(1/2( 1+i))/(1/4 + 1/4)

(1/2( 1+i))/(2/4)

(1/2( 1+i))/(1/2)

1+i =>a+bi

in polar form:

r=sqrt(a^2+b^2)

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

r*cos(θ)=a =>sqrt(2)*cos(θ)=1

=>cos(θ)=1/sqrt(2)

θ=cos^-1(1/sqrt(2))

θ=π/4 (result in radians)

r*sin(θ)=b =>sqrt(2)*sin(θ)=1

=>sin(θ)=1/sqrt(2)

θ=sin^-1(1/sqrt(2))
θ=π/4

sqrt(2) (cos(π/4)+i* sin(π/4)) or sqrt(2) e^((i π)/4)

I can't wait to be at this level of math. I hope Humboldt will at least thank you for this reply. This site needs a major renovation. Where is the owner? Where are the tutors?