# Geometric rotation of (a + bi)

1. Dec 17, 2005

Question, in order to produce a 270 degree geometric rotation of the complex number (a + bi), would this be correct:
(a + bi) * (-i)
It seems logical since a 90 degree rotation results from (a + bi) * (i)
Next question. What would be the equations for rotation of (a + bi) by 45 degrees, 135 degrees, 225 degrees ?
Thanks for help.

2. Dec 17, 2005

### hypermorphism

3. Dec 17, 2005

Is it standard here in the math section of the forum to answer a question with a question ?

4. Dec 17, 2005

### devious_

You can either think of complex numbers as vectors and hence use a suitable transformation matrix, or you can think about their arguments (i.e. the angles they make with the +ve real line).

So, yes, multiplication by -i will rotate a complex number by 270 deg (in the anticlockwise direction), since the argument of -i is 270, and when you multiply complex numbers you add their arguments. This is what hypermorphism was hinting at: Euler's formula can prove this. Since if z=a+ib and w=c+id are two complex numbers with arguments p and q, then z=|z|e^(ip) and w=|w|e^(iq), and hence zw=|wz|e^(i(p+q)).

So in the spirit of the standard of answering a question with a question:
Can you see how to apply this to rotations of any degree?

5. Dec 19, 2005

As devious_ mentioned (and what follows from Euler), the argument of the product of two complex numbers is the sum of their arguments*. So if you want to rotate by x degrees, you'd want to multiply by a number whose argument is x degrees. The argument of $$e^{ix}$$ is x radians*, and it's absolute value is 1 (the absolute value needs to be 1 if you just want a rotation). Convert degrees to radians, and use Euler's if you want to get in a+bi form.