Rotations in Complex Plane

  1. Sep 30, 2008 #1
    I'm reading this book on modern geometry and I was wondering if I'm doing these problems right:

    if I'm give a point 2+i and I'm suppose to rotate is 90 degrees

    first I move it to the origin


    second, I rotate it


    I'm not sure how to interpret that algebraically

    then i replace it

    T^-1(z)= z+(2+i)

    Am I actually doing this right, the book I'm reading is kind of old and doesn't have many worked examples.
  2. jcsd
  3. Sep 30, 2008 #2
    A 90 degree rotation is accomplished by multiplication by i.

    (2+i)i = 2i -1
  4. Sep 30, 2008 #3
    then how is a 45 degree rotation accomplished, in the one example (ill type the whole thing out)

    rotate by 45 degrees at point i

    g(z)=e^(i*pi/4)z= (1+i)z/sqrt(2)
    f^-1(z)=(1+i)(z-i)/sqrt(2) + i

    which equals

    Last edited: Sep 30, 2008
  5. Sep 30, 2008 #4
    I don't understand your notation.

    what is f(z) z-i ?

    To rotate 45 degrees multiply by e^(i*theta), where theta is in radians.

    45 degrees is equal to pi/4 radians.
  6. Sep 30, 2008 #5
    right, i don't understand how the book's example came out with sqrt(2) at the bottom
  7. Sep 30, 2008 #6
    oh sorry, i realized that i forgot equal signs
  8. Oct 3, 2008 #7
    You're notation is still hard to follow. For instance, the letter z is usually used to express a complex number. z = x+iy.

    There are some basic tools you need to perform operations on complex numbers.

    1 Euler's Equation. [tex]\ e^{i \theta} = cos(\theta) + i sin(\theta) [/tex]

    Where [tex]x=cos(\theta)[/tex] and [tex]y= sin(\theta)[/tex], a number in the form [tex]X+iY[/tex] can be expressed in the form [tex]\ Z e^{i \Theta}[/tex].

    (In this case 'Z' is a magnitude, a real positive value--so much for conventions.)

    X,Y,Z, and theta are all real valued numbers, and Z is positive.

    2 Complex Conjugation.

    The complex conjugate of [tex]\ X+iY[/tex] is [tex]\ X-iY[/tex].

    You just negate the imaginary part to get the complex conjugate.

    3 Division.

    [tex] c = a+ib [/tex]
    [tex] z = x+iy [/tex]

    What is the value of c/z expressed in the form X+iY ?

    [tex]\frac{c}{z} = \frac{a+ib}{x+iy} [/tex]

    Multiply the numerator and denominator by the complex conjugate of the denominator.

    [tex]\frac{c}{z} = \frac{(a+ib)(x-iy)}{(x+iy)(x-iy)}[/tex]

    [tex]\ \ \ \ \ \ = \frac{(a+ib)(x-iy)}{x^2 + y^2}[/tex]

    [tex]\ \ \ \ \ \ = \frac{(ax+by) + i(bx - ay)}{x^2 + y^2}[/tex]

    [tex]\ \ \ \ \ \ = \frac{ax+by}{x^2 + y^2} + i \frac{(bx - ay)}{x^2 + y^2}[/tex]
    Last edited: Oct 3, 2008
  9. Oct 3, 2008 #8


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    I'm confused as to what you mean by "rotating a point". Do you mean rotate around the origin? If you mean "rotate the point 2+ i 90 degrees about the origin", you don't need a formula for a general rotation. Rotating the x-axis 90 degrees takes it into the positive y-axis. Rotating the positive y-axis 90 degrees takes it into the negative x-axis. That is, the point (x,y) is rotated into the point (-y, x).
  10. Oct 4, 2008 #9
    It sounds as though you're trying to rotate the complex plane around the point 2+i, rather than rotating the point 2+i around the origin. In this case you're doing the right thing: Given a complex number z, you first translate so that 2+i is at the origin (ie subtract 2+i) then you rotate by 90 degrees (ie multiply by i) and finally you translate back so that the point 2+i is back where it started. Step-by-step:

    z -> z - (2+i)
    z -> iz
    z -> z + (2+i)

    so if you combine all of these into a single mapping you get

    z -> iz + 3 - i

    You can check that plugging 2+i into this formula just gives you 2+i back.

    If you wanted to rotate by an arbitrary angle theta, then you replace step 2 by

    z -> exp(i*theta) z
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