Quantcast Rotations in Complex Plane Text - Physics Forums Library

PDA

View Full Version : Rotations in Complex Plane


chaotixmonjuish
Sep30-08, 09:49 PM
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

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

second, I rotate it

e^(pi/2*i)*z

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.

Phrak
Sep30-08, 10:43 PM
A 90 degree rotation is accomplished by multiplication by i.

(2+i)i = 2i -1

chaotixmonjuish
Sep30-08, 11:03 PM
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

f(z)=z-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

(1+i)z+i*sqrt(2)-i+1/sqrt(2)

Phrak
Sep30-08, 11:09 PM
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.

chaotixmonjuish
Sep30-08, 11:17 PM
right, i don't understand how the book's example came out with sqrt(2) at the bottom

chaotixmonjuish
Sep30-08, 11:28 PM
oh sorry, i realized that i forgot equal signs

Phrak
Oct3-08, 02:12 AM
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. \ e^{i \theta} = cos(\theta) + i sin(\theta)

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

(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 \ X+iY is \ X-iY.

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

3 Division.

c = a+ib
z = x+iy

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

\frac{c}{z} = \frac{a+ib}{x+iy}

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

\frac{c}{z} = \frac{(a+ib)(x-iy)}{(x+iy)(x-iy)}

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

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

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

HallsofIvy
Oct3-08, 01:58 PM
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).

Cexy
Oct4-08, 07:21 AM
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