How Do You Calculate the Modulus and Argument of a Complex Number?

joelstacey
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Homework Statement
For each expression, evaluate its real part, imaginary part, modulus and argument.
1. (e^(i*theta))^2
Relevant Equations
r*e^(i*theta)= r*(cos(theta) + i*sin(theta))
(e^(i*theta))^2 = (sin(theta)+i*cos(theta))^2 = cos(theta)^2 - sin(theta)^2 + 2*i*sin(theta)*cos(theta), so the real part would be: cos(theta)^2 - sin(theta)^2, and the imaginary part would be: 2*i*sin(theta)*cos(theta). But then I don't know where to start with the modulus or the argument?
 
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What are the definition of the modulus and argument of a complex number?
 
Office_Shredder said:
What are the definition of the modulus and argument of a complex number?
I know the modulus is the length of the vector in an argand diagram, and the modulus is the angle it makes with the x axis, but since it is squared i don't see how it works as an argand diagram.
 
You can also write ##(e^{i\theta})^2## as ##e^{2i\theta}##, using the properties of exponents.
Rewriting the above using Euler's formula, we have ##e^{2i\theta} = \cos(2\theta) + i\sin(2\theta)##, which agrees with what you found for the real and imaginary parts (after using double angle identities).

If you have a complex number ##z = a + ib##, how do you find the modulus? From the above, the argument (arg) should be simple to find.
 
I guess a couple of things
1.) Given the real and complex parts, you could write down a new polar coordinates form. The modulus is not hard to compute this way, though I will admit the argument requires knowing some trig trickery.
2.) There's a much simpler way to do this.
## (e^{i\theta})^2 = e^{i\theta} e^{i\theta} = e^{i\theta+i\theta}##.

Note ##(e^a)^b \neq e^{ab}## in general, but when b is an integer you can do this.
 
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joelstacey said:
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so the real part would be: cos(theta)^2 - sin(theta)^2, and the imaginary part would be: 2*i*sin(theta)*cos(theta). But then I don't know where to start with the modulus or the argument?
Do you know the trig identities for ##\cos(2\theta)## and ##\sin(2\theta) ## ?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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