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Find the analytical function of the complex variable $z$ whose imaginary part is : $$v(x,y)= \ln |z|$$

So it's obvious that I'm going to use Cauchy Riemann here, but I feel like there's something else that is missing... So, I have $v(x,y)= \ln\sqrt{x^2+y^2}$ and I find $\partial v/\partial x$, but then what?

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Use Cauchy-Riemann conditions. –  user90036 Aug 12 '13 at 18:14
    
If $v(x,y) = \ln|z|$, then your writing $v(x,y) = \sqrt{x^2+y^2}$ is wrong; it should be $v(x,y)=\ln(\sqrt{x^2+y^2}) = \frac12\ln(x^2+y^2)$... –  Steven Stadnicki Aug 12 '13 at 18:15
    
@njguliyev that is known but how is the question! –  needtostudy Aug 12 '13 at 18:19
    
@StevenStadnicki thank you I corrected it :) –  needtostudy Aug 12 '13 at 18:19
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2 Answers 2

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Even though in principle the Cauchy-Riemann equations can function as differential equations that will allow you to reconstruct a function in this way, in practice doing this directly can be extraordinarily difficult. Exercises of this kind are almost always meant to be solved using a combination of inspired guesswork and previous knowledge.

If you know of the complex exponential and logarithm (and if you don't you certainly have some work cut out for you here), you should immediately notice that the imaginary part you're expected to find here is exactly the real part of the complex logarithm. So we can very quickly manufacture a function that satisfies it, simply by multiplying by $i$: $$ f(z) = i\operatorname{Log} z $$

The only problem with this is that it is not continuous (and thus not analytic) along the cut, wherever your definition of Log places the principal cut. So the next step would be to figure out whether this is unavoidable.

It turns out that this problem cannot be avoided. Clearly if we're given the imaginary part of the function, we can add any real number we want to the entire function without losing analyticity -- so we can arbitrarily decide that the real part of $f(1)$ is going to be $0$.

But then the Cauchy-Riemann equations immediately tell that the real part of $f(x)$ for any positive real $x$ must be $0$ to, simply because $\frac{du}{dx}=\pm \frac{dv}{dy}$ (I don't even bother to look up the sign), and $\frac{dv}{dy}$ is clearly $0$ everywhere on the real axis, given the specified for $v$.

Thus, we now know the entire complex value of $f$ everywhere on the positive real axis. And we can then use a theorem that says that knowing the value of an analytic function on just a small (nontrivial) piece of curve going through its domain determines its values everywhere. So since $i \operatorname{Log} z$ matches the specified values on the positive real axis, it has to match everywhere, at least if the domain of $f$ includes the domain of $\operatorname{Log}$.

Otherwise the domain of $f$ can be stranger, and we then know only that $f(z)$ must be $i$ times some logarithm of $z$ at each point -- but we still cannot make an $f$ that's continuous all the way around the origin, because the argument (which must become minus the real part of $f$) cannot match up.

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This is very nice,danish Robin Williams. –  needtostudy Aug 12 '13 at 19:17
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We know what $v(x,y)$ looks like, but we need $u(x,y)$. The Cauchy Riemann equations tell us that $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\\ \frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x} $$ Can we use these to deduce what $u$ is? Sure! By the first equation, we have $$ u=\int\frac{\partial v}{\partial y}dx+f(y) $$ Sub this into the second equation to get $$ \frac{\partial}{\partial y} \left[\int\frac{\partial v}{\partial y}dx+f(y)\right] = -\frac{\partial v}{\partial x} $$ Use this to find $f(y)$.

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Thank you Om ^_^ –  needtostudy Aug 12 '13 at 18:29
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