Let $u(z)=\arg(z)$ in $D=\mathbb{C} \setminus\mathbb{R}^-$ where $\mathbb{R}^-=(-\infty,0]$. Find Harmonic conjugate for $u$.

Any ideas / hints?

  • $\begingroup$ Hint: Consider the cartesian form of the complex logarithm. $\endgroup$ Aug 6 '17 at 19:22
  • $\begingroup$ @MundronSchmidt As this question is in the material before any mentions of the logarithm,any other ideas? $\endgroup$
    – ChikChak
    Aug 6 '17 at 19:26
  • 1
    $\begingroup$ $u=\arctan\dfrac{y}{x}$ and use Cauchy-Riemann equations. $\endgroup$
    – Nosrati
    Aug 6 '17 at 19:29
  • $\begingroup$ @MyGlasses Thank you. $\endgroup$
    – ChikChak
    Aug 6 '17 at 20:34

I think I might have a simpler solution that requires absolutely no knowledge of complex logarithms.

If we use polar coordinates, then we get $u(r,\theta)=\theta$. Now, we want to find a function $v(r,\theta)$ so that $u(r,\theta) + iv(r,\theta)$ will be holomorphic. Meaning, we have to make sure $u$ and $v$ will satisfy the polar version of the Cauchy-Riemann equations: $$\frac{\partial u}{\partial r}=\frac 1r\frac{\partial v}{\partial \theta},\frac{\partial v}{\partial r}=-\frac 1r\frac{\partial u}{\partial \theta}$$ In our case these translate into the following: $$\frac{\partial v}{\partial \theta}=0,\frac{\partial v}{\partial r}=-\frac 1r$$ This means that $v$ is only a function of $r$ which we can solve for using simple single-real-variable integration to get: $$v(r,\theta)=-\ln r$$


I found it awkward to present the following argument without at least some reference to $\ln z$; so I essentially "invented" it here at equation (4), then developed a couple of basic properties such as $e^{\ln z} = z$. But it doesn't take much "logarithm theory" to answer this question.

In $D = \Bbb C \setminus \Bbb R^- = \Bbb C \setminus (-\infty, 0]$, we may without ambiguity take

$z = re^{i\theta}, \tag 1$

since $\theta$ is restricted to the interval $(-\pi, \pi)$. Furthermore, since the point $r = 0$ is excluded from $D$ as well, $\ln r$ is a well-defined real function on $D$. We may therefore define the complex-valued function on $D$

$g(z) = \ln r + i\theta; \tag 2$

we note that

$e^{g(z)} = e^{\ln r + i\theta} = e^{\ln r} e^{i \theta} = re^{i\theta} = z. \tag 3$

Next, we observe that $z^{-1}$ is holomorphic in $D$, and we define the function $\ln z$ in $D$ via the formula

$\ln z = \displaystyle \int_1^z s^{-1} ds, \tag 4$

where the integral is taken over any path in $D$ which joins $1$ and $z \in D$; since $z^{-1}$ is holomorphic the integral is path-independent, and $\ln z$ is holomorphic in $D$. We show $e^{\ln z} = z$. Consider the holomorphic function

$F(z) = z^{-1} e^{\ln z}; \tag 5$


$\ln 1 = \displaystyle \int_1^1 s^{-1} ds = 0, \tag 6$

we have

$F(1) = 1; \tag 7$


$F'(z) = -z^{-2}e^{\ln z} + z^{-1}(\ln z)' e^{\ln z} = -z^{-2}e^{\ln z} + z^{-2}e^{\ln z} = 0, \tag 8$

since $(\ln z)' = z^{-1}$ from (4). I fowllows fro (8) that $F(z)$ is a constant and so by (7) we see that

$z^{-1}e^{\ln z} = F(z) = 1 \tag 9$

for all $z \in D$; hence

$e^{\ln z} = z. \tag {10}$

We combine (2) and (10):

$e^{\ln z} = e^{g(z)}, \tag {11}$


$e^{(g(z) - \ln z)} = 1; \tag {12}$

if follows that

$g(z) = \ln z + 2n\pi i \tag{13}$

for some $n \in \Bbb Z$; but

$g(1) = 0 = \ln 1, \tag{14}$

so $n = 0$ and

$\ln r + i \theta = g(z) = \ln z \tag{15}$

in $D$. (15) shows that $\ln r + i \theta$ is holomorphic in $D$ and thus the harmonic conjugate of $\theta = \arg(z)$ in $D$ is $\ln r$.


Let the function be $$\arctan\frac yx+i v(x,y).$$

The Cauchy-Riemann equations read

$$v'_y(x,y)=-\frac y{x^2+y^2},\\ v'_x(x,y)=-\frac x{x^2+y^2}.$$

If we integrate the first on $y$,


We now plug it in the second,

$$-\frac x{\sqrt{x^2+y^2}}+c'(x)=-\frac x{x^2+y^2}.$$ This shows that $c(x)$ must be a constant.


$$f(z)=\arctan\frac yx-i\log\sqrt{x^2+y^2}+ic,$$ which you can recognize to be $$-i\log z+ic.$$


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