examples of Cauchy-Riemann equations

To illustrate the Cauchy-Riemann equations, we may consider a fewexamples. Let $f$ be the squaring function, i.e. for any complexnumber $z$, we have $f(z)=z^2$.

We now separate real from imaginary parts. Letting $x$ and $y$ bereal variables we have

$$f(x+iy)={(x+iy)}^2=x^2+2ixy-y^2=(x^2-y^2)+i(2xy).$$

Defining real functions $u$ and $v$ by $f(x+iy)=u(x,y)+iv(x,y)$and taking derivatives, we have

$u(x,y)$$=x^2-y^2$

$v(x,y)$$=2xy$

${\displaystyle \frac{\partial u}{\partial x}}$$=2x$

${\displaystyle \frac{\partial u}{\partial y}}$$=-2y$

${\displaystyle \frac{\partial v}{\partial x}}$$=2y$

${\displaystyle \frac{\partial v}{\partial y}}$$=2x$

Since $\partial u/\partial x=\partial v/\partial y$ and$\partial u/\partial y=-\partial v/\partial x$. theCauchy-Riemann relations are seen to be satisfied.

Next, consider the complex conjugation function. In thiscase, $\overline{x+iy}=x-iy$, so we have $u(x,y)=x$and $v(x,y)=y$. Taking derivatives,

${\displaystyle \frac{\partial u}{\partial x}}$$=1$

${\displaystyle \frac{\partial u}{\partial y}}$$=0$

${\displaystyle \frac{\partial v}{\partial x}}$$=0$

${\displaystyle \frac{\partial v}{\partial y}}$$=-1$

Because $\partial u/\partial x\ne \partial v/\partial y$,the Cauchy-Riemann equations are not satisfied do theconjugation function is not holomorphic. Likewise, one can showthat the functions $\mathrm{\Re },\mathrm{\Im },and|\cdot |$ which appear incomplex analysis are not holomorphic.

For our next example, we try a polynomial. Let $f(Z)=z^3+2z+5$. Writing $z=x+iy$ and $f=u+iv$,we find that $u(x,y)=x^3-3x{y}^2+2x+5$ and$v(x,y)=3x^{2}y-y^3+2y$. Taking partial derivatives,one can confirm that the Cauchy-Riemann equationsare satisfied, so we have a holomorphic function.

More generally, we can show that all complexpolynomials are holomorphic. Since the Cauchy-Riemannequations are linear, it suffices to check that integerpowers are holomorphic. We can do this by an inductionargument. That $f(z)=z$ satisfies the equations istrivial and we have shown that $f(z)=z^2$ alsosatisfies them. Let us assume that $f(z)=z^n$happens to satisfy the Cauchy-Riemann equations for aparticular value of $n$ and write

${(x+iy)}^n$$=u(x,y)+iv(x,y)$

${(x+iy)}^{n+1}$$=\tilde{u}(x,y)+i\tilde{v}(x,y)$

By elementary algebra, we have

$\tilde{u}(x,y)$$=xu(x,y)-yv(x,y)$

$\tilde{v}(x,y)$$=yu(x,y)+xv(x,y).$

By elementary calculus, we have

${\displaystyle \frac{\partial \tilde{u}}{\partial x}}$$=u+x{\displaystyle \frac{\partial u}{\partial x}}-y{\displaystyle \frac{\partial v}{\partial x}}$

${\displaystyle \frac{\partial \tilde{u}}{\partial y}}$$=-v+x{\displaystyle \frac{\partial u}{\partial y}}-y{\displaystyle \frac{\partial v}{\partial y}}$

${\displaystyle \frac{\partial \tilde{v}}{\partial x}}$$=v+y{\displaystyle \frac{\partial u}{\partial x}}+x{\displaystyle \frac{\partial v}{\partial x}}$

${\displaystyle \frac{\partial \tilde{v}}{\partial y}}$$=u+y{\displaystyle \frac{\partial u}{\partial y}}+x{\displaystyle \frac{\partial v}{\partial y}}$

so

${\displaystyle \frac{\partial \tilde{u}}{\partial x}}-{\displaystyle \frac{\partial \tilde{v}}{\partial y}}$$=x\left({\displaystyle \frac{\partial u}{\partial x}}-{\displaystyle \frac{\partial v}{\partial y}}\right)+y\left({\displaystyle \frac{\partial u}{\partial y}}+{\displaystyle \frac{\partial v}{\partial x}}\right)$

${\displaystyle \frac{\partial \tilde{u}}{\partial y}}+{\displaystyle \frac{\partial \tilde{v}}{\partial x}}$$=x\left({\displaystyle \frac{\partial u}{\partial y}}+{\displaystyle \frac{\partial v}{\partial x}}\right)+y\left({\displaystyle \frac{\partial u}{\partial x}}-{\displaystyle \frac{\partial v}{\partial y}}\right)$

Since the terms in parentheses are zero on account of$u$ and $v$ satisfying the Cauchy-Riemann equations,it follows that $\tilde{u}$ and $\tilde{v}$ alsosatisfy the Cauchy-Riemann equations. By induction,$f(z)=z^n$ is holomorphic for all positive integers $n$.

As our next example, we consider the complexsquare root. As shown in the entry taking squareroot algebraically, we have the following equality:

$$\sqrt{x+iy}=\sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}+(\text{sign}y)i\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}$$

Differentiating and simplifying,

$u(x,y)$$=\sqrt{{\displaystyle \frac{\sqrt{x^2+y^2}+x}{2}}}$

$v(x,y)$$=(\text{sign}y)\sqrt{{\displaystyle \frac{\sqrt{x^2+y^2}-x}{2}}}$

${\displaystyle \frac{\partial u}{\partial x}}$$={\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{\frac{x}{\sqrt{x^2+y^2}}+1}{\sqrt{\sqrt{x^2+y^2}+x}}}={\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{\sqrt{\sqrt{x^2+y^2}+x}}{\sqrt{x^2+y^2}}}$

${\displaystyle \frac{\partial u}{\partial y}}$$=(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{\frac{y}{\sqrt{x^2+y^2}}}{\sqrt{\sqrt{x^2+y^2}+x}}}=(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{y}{\sqrt{x^2+y^2}\sqrt{\sqrt{x^2+y^2}+x}}}$

${\displaystyle \frac{\partial v}{\partial x}}$$=(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{\frac{x}{\sqrt{x^2+y^2}}-1}{\sqrt{\sqrt{x^2+y^2}-x}}}=-(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{\sqrt{\sqrt{x^2+y^2}-x}}{\sqrt{x^2+y^2}}}$

${\displaystyle \frac{\partial v}{\partial y}}$$={\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{\frac{y}{\sqrt{x^2+y^2}}}{\sqrt{\sqrt{x^2+y^2}-x}}}={\displaystyle \frac{\sqrt{2}}{4}}{\displaystyle \frac{y}{\sqrt{x^2+y^2}\sqrt{\sqrt{x^2+y^2}-x}}}.$

Pulling out a common factor and placing over a common denominator,

${\displaystyle \frac{\partial u}{\partial x}}-{\displaystyle \frac{\partial v}{\partial y}}$$={\displaystyle \frac{\sqrt{2}}{4\sqrt{x^2+y^2}}}\left(\sqrt{\sqrt{x^2+y^2}+x}-{\displaystyle \frac{y}{\sqrt{\sqrt{x^2+y^2}-x}}}\right)$

$={\displaystyle \frac{\sqrt{2}}{4\sqrt{x^2+y^2}}}{\displaystyle \frac{\sqrt{\sqrt{x^2+y^2}+x}\cdot \sqrt{\sqrt{x^2+y^2}-x}-y}{\sqrt{\sqrt{x^2+y^2}-x}}}$

$={\displaystyle \frac{\sqrt{2}}{4\sqrt{x^2+y^2}}}{\displaystyle \frac{\sqrt{(x^2+y^2)-y^2}-y}{\sqrt{\sqrt{x^2+y^2}-x}}}=0$

${\displaystyle \frac{\partial u}{\partial y}}+{\displaystyle \frac{\partial v}{\partial x}}$$=(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4\sqrt{x^2+y^2}}}\left({\displaystyle \frac{y}{\sqrt{\sqrt{x^2+y^2}+x}}}-\sqrt{\sqrt{x^2+y^2}-x}\right)$

$=(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4\sqrt{x^2+y^2}}}{\displaystyle \frac{y-\sqrt{\sqrt{x^2+y^2}+x}\cdot \sqrt{\sqrt{x^2+y^2}-x}}{\sqrt{\sqrt{x^2+y^2}+x}}}$

$=(\text{sign}y){\displaystyle \frac{\sqrt{2}}{4\sqrt{x^2+y^2}}}{\displaystyle \frac{\sqrt{(x^2+y^2)-x^2}-y}{\sqrt{\sqrt{x^2+y^2}+x}}}=0,$

so the Cauchy-Riemann equations are satisfied. Moregenerally, it can be shown that all complex algebraicfunctions and fractional powers satisfy theCauchy-Riemann equations. However, as suggested bythe above derivation, a direct verification couldbe tedious, so it is better to use an indirect approach.

Finally, we finish up with two examples of transcendentalfunctions, the complex exponential and the complex logarithm,The complex exponential is defined as$\exp (x+iy)=\exp (x)(\cos y+i\sin y)$. Hencewe have

$u(x,y)$$=\exp (x)\cos (y)$

$v(x,y)$$=\exp (x)\sin (y)$

${\displaystyle \frac{\partial u}{\partial x}}$$=\exp (x)\cos (y)$

${\displaystyle \frac{\partial u}{\partial y}}$$=-\exp (x)\sin (y)$

${\displaystyle \frac{\partial v}{\partial x}}$$=\exp (x)\sin (y)$

${\displaystyle \frac{\partial v}{\partial y}}$$=\exp (x)\cos (y)$

Thus we see that the complex exponential function is holomorphic.

The complex logarithm may be defined as $\log (x+iy)=1/2\log (x^2+y^2)+i\arctan (y/x)$. Hence we have

$u(x,y)$$=1/2\log (x^2+y^2)$

$v(x,y)$$=\arctan (y/x)$

${\displaystyle \frac{\partial u}{\partial x}}$$={\displaystyle \frac{x}{x^2+y^2}}$

${\displaystyle \frac{\partial u}{\partial y}}$$={\displaystyle \frac{y}{x^2+y^2}}$

${\displaystyle \frac{\partial v}{\partial x}}$$={\displaystyle \frac{-y/x^2}{1+{(y/x)}^2}}=-{\displaystyle \frac{y}{x^2+y^2}}$

${\displaystyle \frac{\partial v}{\partial y}}$$={\displaystyle \frac{1/x}{1+{(y/x)}^2}}={\displaystyle \frac{x}{x^2+y^2}}$

Hence the complex logarithm is holomorphic.