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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.