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Rotations in Complex Plane 
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#1
Sep3008, 08:49 PM

P: 287

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. 


#2
Sep3008, 09:43 PM

P: 4,512

A 90 degree rotation is accomplished by multiplication by i.
(2+i)i = 2i 1 


#3
Sep3008, 10:03 PM

P: 287

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


#4
Sep3008, 10:09 PM

P: 4,512

Rotations in Complex Plane
I don't understand your notation.
what is f(z) zi ? To rotate 45 degrees multiply by e^(i*theta), where theta is in radians. 45 degrees is equal to pi/4 radians. 


#5
Sep3008, 10:17 PM

P: 287

right, i don't understand how the book's example came out with sqrt(2) at the bottom



#6
Sep3008, 10:28 PM

P: 287

oh sorry, i realized that i forgot equal signs



#7
Oct308, 01:12 AM

P: 4,512

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 valueso 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]\ XiY[/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)(xiy)}{(x+iy)(xiy)}[/tex] [tex]\ \ \ \ \ \ = \frac{(a+ib)(xiy)}{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] 


#8
Oct308, 12:58 PM

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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 xaxis 90 degrees takes it into the positive yaxis. Rotating the positive yaxis 90 degrees takes it into the negative xaxis. That is, the point (x,y) is rotated into the point (y, x).



#9
Oct408, 06:21 AM

P: 70

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. Stepbystep:
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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