Euler's formula and differential equations

  • Thread starter Referos
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  • #1
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Homework Statement



Consider the complex number [tex]z=rcos\theta + irsin\theta[/tex], where [tex]r=|z|[/tex], [tex]\theta=arg(z)[/tex] ([tex]r[/tex] is constant, [tex]\theta[/tex] is a variable). Show that [tex]\frac{dz}{d\theta}=iz[/tex], then solve this differential equation to show that [tex]z=re^{i\theta}[/tex].


Homework Equations



[tex]z=rcos\theta + irsin\theta[/tex]

[tex]z=re^{i\theta}[/tex]

[tex]r=|z|[/tex]

[tex]\theta=arg(z)[/tex]

The Attempt at a Solution


I had no problem showing that [tex]\frac{dz}{d\theta}=iz[/tex]. My problem was with the second part; I thought it was pretty straightforward, but it simply didn't work.

[tex]\frac{dz}{d\theta}=iz[/tex]

[tex]\frac{d\theta}{dz}=\frac{1}{iz}[/tex]

[tex]i\int\frac{d\theta}{dz}dz=\int\frac{1}{z}dz[/tex]

[tex]i\theta=ln|z|+C[/tex]

[tex]i\theta-C=lnr[/tex]

[tex]e^{i\theta-C}=r[/tex]

[tex]e^{-C}\times e^{i\theta} = r[/tex]

Which is impossible! For this to be Euler's formula, [tex]e^{-C}[/tex] needs to be equal to [tex]r[/tex], which I suppose is okay because [tex]e[/tex], [tex]C[/tex] and [tex]r[/tex] are constants. But the problem is the [tex]r[/tex] in the right-hand side of the equation. The only way I see for it to be [tex]z[/tex] instead of [tex]r[/tex] would be if [tex]\int\frac{1}{z}dz=lnz[/tex] rather than [tex]\int\frac{1}{z}dz=ln|z|[/tex]. But this is cannot be, or can it?
Thanks.
 

Answers and Replies

  • #2
34,542
6,248

Homework Statement



Consider the complex number [tex]z=rcos\theta + irsin\theta[/tex], where [tex]r=|z|[/tex], [tex]\theta=arg(z)[/tex] ([tex]r[/tex] is constant, [tex]\theta[/tex] is a variable). Show that [tex]\frac{dz}{d\theta}=iz[/tex], then solve this differential equation to show that [tex]z=re^{i\theta}[/tex].


Homework Equations



[tex]z=rcos\theta + irsin\theta[/tex]

[tex]z=re^{i\theta}[/tex]

[tex]r=|z|[/tex]

[tex]\theta=arg(z)[/tex]

The Attempt at a Solution


I had no problem showing that [tex]\frac{dz}{d\theta}=iz[/tex]. My problem was with the second part; I thought it was pretty straightforward, but it simply didn't work.

[tex]\frac{dz}{d\theta}=iz[/tex]
It's quicker to move things around to get z and dz on one size and [itex]d\theta[/itex] on the other.
[tex]\frac{dz}{z }=id\theta[/tex]
Now integrate to get an equation involving z and [itex]\theta[/itex].
[tex]\frac{d\theta}{dz}=\frac{1}{iz}[/tex]

[tex]i\int\frac{d\theta}{dz}dz=\int\frac{1}{z}dz[/tex]

[tex]i\theta=ln|z|+C[/tex]
What if you work this through without replacing |z| as you did in the following equation?
[tex]i\theta-C=lnr[/tex]

[tex]e^{i\theta-C}=r[/tex]

[tex]e^{-C}\times e^{i\theta} = r[/tex]

Which is impossible! For this to be Euler's formula, [tex]e^{-C}[/tex] needs to be equal to [tex]r[/tex], which I suppose is okay because [tex]e[/tex], [tex]C[/tex] and [tex]r[/tex] are constants. But the problem is the [tex]r[/tex] in the right-hand side of the equation. The only way I see for it to be [tex]z[/tex] instead of [tex]r[/tex] would be if [tex]\int\frac{1}{z}dz=lnz[/tex] rather than [tex]\int\frac{1}{z}dz=ln|z|[/tex]. But this is cannot be, or can it?
Thanks.
 
  • #3
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What if you work this through without replacing |z| as you did in the following equation?
Thanks for the help, but I don't see how not replacing |z| by r solves anything. I would still get a |z| instead of a z. Besides, if I have defined |z| to be r, then I should be able to freely substitute |z| by r without any contradiction, shouldn't I?
 

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