Solve Complex Numbers & Euler's Formula Problems - 65 Characters

  • Thread starter Thread starter SammC
  • Start date Start date
  • Tags Tags
    Formula
SammC
Messages
17
Reaction score
0
Theres a couple problems I am working on that involve complex numbers or euler's formula.
e^+-(ix) = cos(x) +- isin(x)

-----------------------
1. A complex number can be written in rectangular coordinates as z = x+ jy. Write the relations to calculate the
polar form, z = (r,theta) or z = re^(j*theta) .

For this one I am more confused about what he's asking or how to show the work... i think
r = sqrt(x^2 + y^2)
and
theta = tan^-1(y/x)

But I'm not really sure if that's what he's looking for


-----------------------
3. Convert cos(wt + f) into the sum of complex exponentials.
Now i know that cos(x) = (e^(ix) + e^-(ix))/2

Is this as simple as replacing x with (wt + f)?


------------------------
5. Compute [(1+ i*sqrt(3))/2]^2 and (1 + j)^4
a) directly (using rectangular representation)
b) using complex exponentials

How do i go about this for both of these approaches, I am not entirely sure how to do either approach.
 
Physics news on Phys.org
For 1 and 3 it seems you are on the right track.

For 5, a) directly using the rectangular representation means just multiplying it out, and b) using complex exponentials means converting the numbers to their complex exponential form and then raising them to the powers.

Evidently 5 is intended as an object lesson in how much nicer it is to exponentiate the complex form compared to the (a+bi) form, even though most students who are first encountering Euler's formula are much more comfortable with the more standard algebraic method because it's just grinding coefficients.
 
Okay.. so I'm a little stuck on 5b...
[1 + i * sqrt(3)/2]^2

would i do:

r = sqrt[1^2 + (sqrt(3)/2)^2]
theta = tan^-1[(sqrt(3)/2)/1]

= [r * e ^ (i * theta)] ^ 2
 
I just tried working 5b through for (1+I)^4:

r = sqrt(1^2 + 1^2) = sqrt(2)
theta = tan^-1(1/1) = pi/4

... so

(1 + i)^4 = [sqrt(2) * e^(i*pi/4)] ^ 4

apply euler's...

= (sqrt(2) * [cos(pi/4) + i * sin(pi/4)])^4

..simplify

= (sqrt(2) * [sqrt(2)/2 + i * sqrt(2)/2])^4

..multiply

= (1 + i)^4

..well... duh.

I think my approach might be wrong.
--------------------------------------
EDIT
figured it out:

(1 + i)^4 = [sqrt(2) * e^(i*pi/4)] ^ 4
= sqrt(2)^4 * e^(4i*pi/4)
= 4 * e^(i*pi)
= 4*-1
=-4
 
Last edited:
One last thing

x1(t) = 5cos(400pi*t +0.5pi)
x2(t) = 5cos(400pi*t -0.25pi)
x3(t) = 5cos(400pi*t +0.4pi)
x4(t) = 5cos(400pi*t - 0.9pi)

I need to express each of those as complex exponentials.. then express the sum.. the only way i know how to do that is to use the fact that
cos(x) = [e^(i*x)+ e^-(i*x)]
and make x = 400pi*t +0.5pi.. etc

which gives me four really long equations.. which seems like it can't possibly be right for what is being asked. At the very least tehre has to be a better way to express the sum of all of them?
 
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...
Back
Top