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

anonymousanonymous
in Science & Mathematics

polar form z=re^iθ

cartesian form: x=iy

'i' is the imaginary number.

just give me the general steps.

if you don't know how to: solve this problem please.

i^(3/2)

thanks!!

Comments

  • zpconnzpconn

    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

  • spinellispinelli

    Convert To Cartesian Form

  • nyphdinmdnyphdinmd

    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)

  • anonymousanonymous

    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

  • jgjg

    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

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