# Complex Analysis - find v given u

• hotvette
In summary, the partial derivatives of u and v are calculated and equated to solve for the unknown function g(y). By setting g'(y) = 0, the final solution for v is obtained as (-y)/(x^2+y^2) + C.
hotvette
Homework Helper
Homework Statement
Let f = u + iv be analytic. Find v for given u.
Relevant Equations
\begin{align*}
u &= \frac{x}{x^2+y^2} = x (x^2+y^2)^{-1} \\
\rule{0mm}{18pt} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}
\qquad\qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
\end{align*}
Solution Attempt:

\begin{align}
\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} = (x^2+y^2)^{-1} -x (x^2+y^2)^{-2} (2x)
= (x^2+y^2)^{-1} - 2x^2 (x^2+y^2)^{-2}
\\
\rule{0mm}{18pt} \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} = -x (x^2+y^2)^{-2} (2y) = -2xy(x^2+y^2)^{-2}
\\
\rule{0mm}{18pt} \frac{\partial v}{\partial x} &= 2xy(x^2+y^2)^{-2}
\\
\rule{0mm}{18pt} v &= -y(x^2+y^2)^{-1} + g(y)
\\
\rule{0mm}{18pt} \frac{\partial v}{\partial x} &= y(x^2+y^2)^{-2} (2x)
= 2xy(x^2+y^2)^{-2} \qquad \checkmark
\\
\rule{0mm}{18pt} \frac{\partial v}{\partial y} &= -(x^2+y^2)^{-1} + 2y^2 (x^2+y^2)^{-2} + g'(y)
\end{align}
Equating (1) and (6):

-(x^2+y^2)^{-1} + 2y^2 (x^2+y^2)^{-2} + g'(y) = (x^2+y^2)^{-1} - 2x^2 (x^2+y^2)^{-2}

I get stuck at the last step. I can't see any function of y that would satisfy the equation. What am I doing wrong?

Last edited:
##g'(y)=0##?

PeroK
I don't know if this is in the spirit of what is expected for solving the problem, but:
##\frac {x}{x^2+y^2}## is the real part of ##\frac {1}{x+iy}## which is analytic except at ##z=0##.
The real and imaginary parts of ##\frac {1}{x+iy}## are easily found by multiplying both numerator and denominator by the conjugate, ##x-iy##.

PeroK
Also, it would have been better if you had simplified:
$$\frac{\partial u}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$$

FactChecker
hotvette said:
I get stuck at the last step. I can't see any function of y that would satisfy the equation. What am I doing wrong?
Oh no! You did everything right and stopped a little too soon. Try ##g(y) \equiv 0##.

PeroK said:
Also, it would have been better if you had simplified:
$$\frac{\partial u}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$$
I agree. It is hard to recognize things as done in the OP.
When taking the derivative of a quotient, I like to use the quotient rule rhyme:
If the quotient rule you wish to know,
It's low D-high less high D-low.
Draw a line and down below
The denominator squared must go.

Drat, I see it now. Thanks!

\begin{align*}
\frac{y^2-x^2}{(x^2+y^2)^2} &= \frac{y^2-x^2}{(x^2+y^2)^2} + g'(y) \\
\rule{0mm}{18pt} g'(y) & = 0 \\
\therefore v &= \frac{-y}{x^2+y^2} + C
\end{align*}

PeroK

## 1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of complex numbers and functions. It involves analyzing the properties and behavior of functions that are defined on the complex plane, which is a two-dimensional space where the horizontal axis represents the real numbers and the vertical axis represents the imaginary numbers.

## 2. What is the purpose of finding v given u in complex analysis?

Finding v given u in complex analysis is an important step in understanding the behavior of complex functions. It allows us to visualize the function in the complex plane and gain insights into its properties, such as continuity, differentiability, and analyticity.

## 3. How is v calculated given u in complex analysis?

In complex analysis, v can be calculated using the Cauchy-Riemann equations, which relate the partial derivatives of a complex function to its real and imaginary parts. The equations are: ∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x. By solving these equations, we can find the expression for v in terms of u.

## 4. Can complex analysis be applied in other fields of science?

Yes, complex analysis has numerous applications in various fields of science, such as physics, engineering, and economics. In physics, it is used to study phenomena involving waves and oscillations, while in engineering it is used to analyze electrical circuits and signal processing. In economics, it is used to model complex systems and make predictions.

## 5. What are some real-life examples of complex analysis?

Some real-life examples of complex analysis include the study of fluid dynamics, where complex functions are used to model the behavior of fluids, and the analysis of electric circuits, where complex numbers are used to represent the voltage and current in AC circuits. Other examples include the study of wave propagation, quantum mechanics, and number theory.

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