Euler's formula and differential equations

[Referos]

Homework Statement

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

Homework Equations

z=rcos\theta + irsin\theta

z=re^{i\theta}

r=|z|

\theta=arg(z)

The Attempt at a Solution

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

\frac{dz}{d\theta}=iz

\frac{d\theta}{dz}=\frac{1}{iz}

i\int\frac{d\theta}{dz}dz=\int\frac{1}{z}dz

i\theta=ln|z|+C

i\theta-C=lnr

e^{i\theta-C}=r

e^{-C}\times e^{i\theta} = r

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

 

[Mark44]

Homework Statement

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

Homework Equations

z=rcos\theta + irsin\theta

z=re^{i\theta}

r=|z|

\theta=arg(z)

The Attempt at a Solution

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

\frac{dz}{d\theta}=iz

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

\frac{d\theta}{dz}=\frac{1}{iz}

i\int\frac{d\theta}{dz}dz=\int\frac{1}{z}dz

i\theta=ln|z|+C

What if you work this through without replacing |z| as you did in the following equation?

i\theta-C=lnr

e^{i\theta-C}=r

e^{-C}\times e^{i\theta} = r

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

 

[Referos]

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?