Complex numbers

1. Sep 17, 2005

Atilla1982

Anyone got a good link to a place that explains complex numbers?

2. Sep 17, 2005

arildno

What do you want to know about them?

3. Sep 17, 2005

Atilla1982

I'm having a hard time rewriting from one form to another, carthesian - polar and so on.

4. Sep 17, 2005

Hurkyl

Staff Emeritus
Well, the procedure is essentially identical to converting between rectangular and polar coordinates on the good ol' real plane, so if that's where you're having trouble, you can pick up one of your old textbooks and review.

5. Sep 17, 2005

arildno

As Hurkyl said, just think of the good ole plane here.

Examples:
Suppose that a complex number z is given by:
z=a+ib
where a,b are real numbers, and i the imaginary unit.
Then, multiply z with 1 in the following manner:
$$z=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}}}(a+ib)={\sqrt{a^{2}+b^{2}}}(\frac{a}{\sqrt{a^{2}+b^{2}}}+i\frac{b}{\sqrt{a^{2}+b^{2}}})$$
Find the angle $$\theta$$ that is the solution of the system of equations:
$$\frac{a} {\sqrt{a^{2}+b^{2}}}=\cos\theta,\frac{b}{\sqrt{a^{2}+b^{2}}}=\sin\theta$$
Thus, defining $$|z|={\sqrt{a^{2}+b^{2}}}$$, we get:
$$z=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}$$
by definition of the complex exponential.

6. Sep 25, 2005

jaap de vries

buy a ti 89 or voyage 200 and your problems are forever solved

7. Sep 25, 2005

Edgardo

8. Sep 28, 2005

Edwin

Euler Identity

Here is a cool trick for calculating pi derived from Euler Identity.

e^(i*(pi/2)) = Cos(90) + i*Sin(90)

ln(e^(i*(pi/2)) = ln(Cos(90) +i*Sin(90))

i*(pi/2)*lne = ln(Cos(90) +i*Sin(90))

pi = (1/i)*(2)*ln(Cos(90) +i*Sin(90))

pi = (i^4/i)*(2)*ln(Cos(90) +i*Sin(90))

pi = (-i)*(2)*ln(Cos(90) +i*Sin(90))

pi = (-2i)*ln(Cos(90) +i*Sin(90))

pi = ln((Cos(90) +i*Sin(90))^(-2i))

pi = ln(1/(Cos(90) +i*Sin(90))^(2i))

Just a cool trick!

Best Regards,

Edwin G. Schasteen

9. Sep 29, 2005

philosophking

"buy a ti 89 or voyage 200 and your problems are forever solved"

if his problem is understanding how certain things work, then i think his problem would stay untouched if he bought one of these caluclators.

10. Sep 29, 2005

Edwin

I have a TI-83 plus and a TI Voyage 200, and I carry them both with me everywhere I go! They are truely amazing computation devices for those of us that are numerically challenged or just plain lazy.

Best Regards,

Edwin

11. Oct 4, 2005