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Complex numbers

by Atilla1982
Tags: complex, numbers
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Atilla1982
#1
Sep17-05, 03:51 AM
P: 18
Anyone got a good link to a place that explains complex numbers?
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arildno
#2
Sep17-05, 06:36 AM
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What do you want to know about them?
Atilla1982
#3
Sep17-05, 07:21 AM
P: 18
I'm having a hard time rewriting from one form to another, carthesian - polar and so on.

Hurkyl
#4
Sep17-05, 08:01 AM
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Complex numbers

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.
arildno
#5
Sep17-05, 08:34 AM
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Quote Quote by Atilla1982
I'm having a hard time rewriting from one form to another, carthesian - polar and so on.
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:
[tex]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}}})[/tex]
Find the angle [tex]\theta[/tex] that is the solution of the system of equations:
[tex]\frac{a} {\sqrt{a^{2}+b^{2}}}=\cos\theta,\frac{b}{\sqrt{a^{2}+b^{2}}}=\sin\theta[/tex]
Thus, defining [tex]|z|={\sqrt{a^{2}+b^{2}}}[/tex], we get:
[tex]z=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}[/tex]
by definition of the complex exponential.
jaap de vries
#6
Sep25-05, 01:42 AM
P: 254
buy a ti 89 or voyage 200 and your problems are forever solved
Edgardo
#7
Sep25-05, 04:31 AM
P: 686
Maybe these websites will be a little help for you:

http://mathworld.wolfram.com/ComplexNumber.html
http://www.clarku.edu/~djoyce/complex/polar.html

I can only give you one tip:
(i) Get familiar with graphical interpretation of the sine and cosine in a circle.
(ii) Really try to understand the formula by examining the drawing of a complex number (like in the link above).

[tex]z=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}[/tex]
Edwin
#8
Sep28-05, 11:03 PM
P: 167
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
philosophking
#9
Sep29-05, 09:44 AM
P: 174
"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.
Edwin
#10
Sep29-05, 03:46 PM
P: 167
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
jaap de vries
#11
Oct4-05, 07:04 PM
P: 254
[QUOTE=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.



Well said my friend. Understanding how it works does not mean that you need to bust your chops doing it the hard way all the time.


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