Define the complex number Z = u^v

[Bruno Tolentino]

If I define the complex number z = r exp(i θ) how z = u^v, so, how to express u and v in terms of r and θ?

u(r, θ) = ?
v(r, θ) = ?

And the inverse too:

r(u, v) = ?
θ(u, v) = ?

 

[SteamKing]

If I define the complex number z = r exp(i θ) how z = u^v, so, how to express u and v in terms of r and θ?

u(r, θ) = ?
v(r, θ) = ?

And the inverse too:

r(u, v) = ?
θ(u, v) = ?

Have you looked at this article?

http://mathworld.wolfram.com/ComplexExponentiation.html

 

[HallsofIvy]

I don't understand your question. You ask, apparently, for z in terms of "u" and "v" but have not said what "u" and "v" are! Are you referring to the representation of a complex function as "z(x+ iy)= u(x,y)+ iv(x,y)" where u and v are real valued function of the real variables x and y? If so then $z= re^{i\theta}$ is NOT "$u^v$". $z= re^{i\theta}= r(cos(\theta)+ isin(\theta)$ so that $u= r cos(\theta)$ and $v= r sin(\theta)$.

 

[mathman]

If I define the complex number z = r exp(i θ) how z = u^v, so, how to express u and v in terms of r and θ?

u(r, θ) = ?
v(r, θ) = ?

And the inverse too:

r(u, v) = ?
θ(u, v) = ?

I'm not sure what you are looking for, but $u=re^{i\theta},\ v=1$ works.
However there are an infinite number of possibilities, by using $u=r^ne^{ni\theta}\ and\ v=\frac{1}{n}$.

 

[Bruno Tolentino]

Z is a complex number, u is a real number and v is a real number too. Is just another way of express the complex numbers...

So, is possible convert the expression z = x + i y in z = u^v ? Is possible express u and v in terms of x and y?

 

[Svein]

So, is possible convert the expression z = x + i y in z = uv ? Is possible express u and v in terms of x and y?

Well, let's try. You know that $z=x+iy$ can also be expressed as $z=re^{i\phi}$, where $r=\sqrt{x^{2}+y^{2}}$ and $\phi = \arcsin(\frac{y}{r})$. Therefore, we obviously have $z=e^{\ln r}\cdot e^{i\phi}= e^{\ln r + i\phi}$. ...

 

[Bruno Tolentino]

Yeah! I thought this... but, I was unsatisfied com this 'conversion' and so I posted my doubt here because the most experiente could see something better...

Anyway! Thank you!