How do you convert a polar form into a cartesian form?

Comments

• zpconn

  May 2021

  You need to use Euler's formula. This states

  r * e^(iθ) = r * ( cos(θ) + i * sin(θ) )

  As for computing i^(3/2), the principal branch is given by

  e^(ln(i) * 3/2)

  = e^( (0 + i * pi/2) * 3/2 )

  = e^(i * 3pi/4)

  = cos(3pi/4) + i * sin(3pi/4)

  = -sqrt(2)/2 + i * sqrt(2)/2
• spinelli

  May 2021

  Convert To Cartesian Form
• nyphdinmd

  May 2021

  Cartesian form is

  z= x + iy

  r = sqrt(x^2 + y^2) q = atan(y/x)

  x = rcos(q) y = rsin(q)

  z = r*(cos(q) + i sin(q)) where q is the angle theta

  now z = i^(3/2) = (0+ i)^(3/2) in cartesian form

  so r = sqrt(0^2 + 1^2) = 1

  q = atan(1/0) =atan(infinity) = pi/2 radians

  z = 1*(e^(i*pi/2))^3/2 = e^(i*3*pi/4)
• anonymous

  May 2021

  r = -2/sinθ y = r sinθ = -2 sinθ/sinθ y = -2 ================== r = (1/cosθ) + 3 x = r cosθ = 1 + 3 cosθ y = r sin θ = tanθ + 3sinθ = (y/x) + 3 sinθ 3 cosθ = x - 1 3 sin θ = y - (y/x) 9 cos²θ + 9 sin²θ = (x - 1)² + [y - (y/x)]² (x - 1)² + [y - (y/x)]² = 9
• jg

  May 2021

  z = re^(θi)

  to find the x use

  x= rsin(θ)

  to find the y use

  y = rcos(θ)

  i^(3/2) = (cos(90) + isin(90))^(3/2)

  = (cos(135) + isin(135))

  = -&radic(2)/2 + i√(2)/2