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Cauchy–Riemann Equations

The Cauchy Riemann equations consist of a system of two partial differential equations which together form a necessary and sufficient condition for a complex function to be complex differentiable that is holomorphic with certain continuity and differentiability criteria.

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Cauchy Riemann Equations Proof

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The Cauchy Riemann equations for a pair of given real valued functions in two variables say, u (x, y) and v (x, y) are the following two equations:

$\frac{\partial u}{\partial x}$=$\frac{\partial v}{\partial y}$     and $\frac{\partial u}{\partial y}$=-$\frac{\partial v}{\partial x}$
 In a typical way, the values ‘u’ and ‘v’ are taken as the real and the imaginary parts of a complex valued function of a single complex variable respectively,
$z = x + i y, g (x + iy)$ = $u (x, y) + i v (x, y)$

If we are given that the functions u and v are differentiable at real values at a point in an open subset of the set of complex numbers that is $C$ which can be taken as functions that are from $R^2$ to $R$. This will imply that the partial derivatives of $u$ and $v$ do exist and thus we can also approximate smaller variations of ‘$g$’ in linear form. Then we say that $g = u + iv$ is differentiable at complex values at that particular point iff the Cauchy Riemann equations at that point are satisfied by the partial derivatives of $u$ and $v$.

Let us assume that $g (x, y)$ = $u (x, y) + i v (x, y)$, where we have,
     $z \equiv  x + i y$

So we have,
    $dz$ = $dx$ + $i dy$

Then the total derivative of ‘$f$’ with respect to ‘$z$’ is:

 $\frac{df}{dz}$=$\frac{\partial f}{\partial x}$ $\times$ $\frac{\partial x}{\partial z}$+ $\frac{\partial f}{\partial y}$ $\times$ $\frac{\partial y}{\partial z}$

or $\frac{df}{dz}$=$\frac{1}{2}$ [$\frac{\partial f}{\partial x}$ - $i$ $\frac{\partial f}{\partial y}$]

The above equation can be written in terms of ‘$u$’ and ‘$v$’ as below:

    $\frac{df}{dz}$=$\frac{1}{2}$[($\frac{\partial u}{\partial x}$+$i$ $\frac{\partial v}{\partial x}$)-$i$($\frac{\partial u}{\partial y}$+$i$ $\frac{\partial v}{\partial y}$)]

or $\frac{df}{dz}$=$\frac{1}{2}$[($\frac{\partial u}{\partial x}$ +$i$ $\frac{\partial v}{\partial x}$)+(-$i$ $\frac{\partial u}{\partial y}$+ $\frac{\partial v}{\partial y}$)]

Along the x – axis or we can say the real axis, $\frac{\partial f}{\partial y}$ = 0. Thus we get,

$\equiv$ $\frac{df}{dz}$=$\frac{1}{2}$[$\frac{\partial u}{\partial x}$-$i$ $\frac{\partial v}{\partial x}$]

Along the y – axis or we can say the imaginary axis, $\frac{\partial f}{\partial x}$ = 0. Thus we have,

 $\frac{df}{dz}$=$\frac{1}{2}$(-$i$ $\frac{\partial u}{\partial y}$+$\frac{\partial v}{\partial y}$)

If we are given that ‘$g$’ is differentiably complex, then we must have the same value of the derivative for a given $dz$, irrespective

of the orientation of it. This implies that the above two equations must be equal to each other, which will further require that,

$\frac{\partial u}{\partial x}$=$\frac{\partial v}{\partial y}$    and   $\frac{\partial v}{\partial x}$=-$\frac{\partial u}{\partial y}$

These two equations are termed as the Cauchy Riemann equations.

These lead to the following conditions:

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

 $\frac{\partial ^{2}v}{\partial x^{2}}$= -$\frac{\partial ^{2}v}{\partial y^{2}}$

We can also write the Cauchy Riemann equations concisely as below:

$\frac{df}{dz}$=$\frac{1}{2}$[$\frac{\partial f}{\partial x}$+$i$ $\frac{\partial f}{\partial y}$]

= $4\frac{1}{2}$[($\frac{\partial u}{\partial x}$+$i$ $\frac{\partial v}{\partial x}$)+($\frac{\partial u}{\partial y}$+$i$ $\frac{\partial v}{\partial y}$)

= $\frac{1}{2}$[($\frac{\partial u}{\partial x}$-$\frac{\partial v}{\partial y}$)+$i$($\frac{\partial u}{\partial y}$-$\frac{\partial v}{\partial x}$)

= 0
where we are given that z-bar is the complex conjugate.

Cauchy Riemann Equations in Polar Coordinates

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Let us assume that ‘g’ is a function of complex values which is differentiable at a certain point say z0 of the complex plane. Here we need to modify the method which resulted in the Cartesian version of the Cauchy Riemann equations.

Let us take $z_0 \neq 0$. We write $z$ = $re^{i\theta }$ and  z0 = $r_{0}e^{i\theta_{0} }$. Now we express the real and imaginary parts of g as the functions in terms of r and $\theta $.
    g ($re^{i\theta }$) = u (r, $\theta $) + i v (r, $\theta $)

Step I : By the definition of differentiability at $z_0$, let us say z $\rightarrow  $  $z_0$ along the ray $\theta $ = $\theta _{0}$. Then, the following limit will exist:

$f'(z_0)$ = $lim_{(r,\theta_0,r_0)}$ $\frac{f(re^{i\theta_0})-f(r_0e^{i\theta_0})}{re^{i\theta_0}-r_0e^{i\theta_0}}$

= $\frac{1}{e^{i\theta_0}}$ $lim_{r\rightarrow r_0}$ $\frac{u(r, \theta_0)-u(r_0, \theta_0)+iv(r, \theta_0)-v(r_0, \theta_0)}{r-r_0}$

=$ e^{-i \theta_0} [lim_{r\rightarrow r_0}$ $\frac{u(r, \theta_0)- u(r_0, \theta_0)}{r-r_0}$ + $i$ $\frac{v(r, \theta_0)- v(r_0, \theta_0)}{r-r_0}$]

Both the limits of the last line will exist. This is because the limit of the first line does exist and a complex function does have a limit at a point only if the real and imaginary parts of it do have limits. Now, these limits are same as the respective partial derivatives of u and v that too with respect to ‘r’, at the polar coordinates (r0, $\theta _{0}$). So we obtain the following result:
$f'( z_0)$ = $e^{-i\theta _{0}}$ [ $\frac{\partial u}{\partial r}$ (r0, $\theta _{0}$) + $i$ $\frac{\partial v}{\partial r}$ ($r_0$, $\theta _{0}$)             (1)

Step II : From the definition of differentiability at z = 0, let z $\rightarrow  $ z0 along the circle r = r0 in such a way that the following holds true:
$f' ( z_0)$ = $lim_{(\theta \rightarrow  \theta _{0})}$ [(f ($r_{0}e^{i\theta }$) – f ($r_{0}e^{i\theta_{0} }$)) / ($r_{0}e^{i\theta_{0} }$ – $r_{0}e^{i\theta_{0} }$)]

= 1/r0  $lim_{(\theta \rightarrow  \theta _{0})}$   [(u (r0,  $\theta $) – u (r0,  $\theta _{0}$) + i [v (r0,  $\theta $) – v (r0,  $\theta _{0}$)]) / ($e^{i\theta }$ – $e^{i\theta_{0} }$)]

= 1/r0$lim_{(\theta \rightarrow  \theta _{0})}$ [(u (r0, $\theta $) – u (r0,  $\theta _{0}$)) / ($\theta $ –  $\theta _{0}$) + i (v (r0, $\theta $) – v (r0,  $\theta _{0}$)) / ($\theta $ –  $\theta _{0}$)] X ($\theta $ –  $\theta _{0}$) / ($e^{i\theta }$ – $e^{i\theta_{0} }$)]
As $\theta $ converges to $\theta _{0}$, the quotients difference in the square brackets will also converge. If they converge at all to the partial derivatives of u and v with respect to $\theta $ that is evaluated at the polar coordinates (r0,  $\theta _{0}$).

This convergence will only happen if we are able to prove that the last fraction is having a limit as $\theta $ $\rightarrow  $ $\theta _{0}$. The reciprocal of the fraction of whose convergence we want to establish is following:

 ($e^{i\theta }$ – $e^{i\theta_{0} }$) / ($\theta $ – $\theta _{0}$) = (cos$\theta $ – cos $\theta _{0}$) / ($\theta $ – $\theta _{0}$) + i (sin $\theta $ – sin $\theta _{0}$) / ($\theta $ – $\theta _{0}$)

which tends to the following, as $\theta $ $\rightarrow  $ $\theta _{0}$,

 [$\frac{d}{d \theta} $ cos $\theta $]$_{(\theta  = \theta _{0})}$ + i [$\frac{d}{d \theta} $ sin $\theta $]$_{(\theta = \theta _{0})}$ = - sin $\theta _{0}$ + i cos $\theta _{0}$ = i e$^{i \theta _{0}}$

When we put everything together we get,

    $f' ( z_0)$ = 1/($i r_0$ $e^{i\theta_{0} }$) [$\frac{\partial u}{\partial \theta }$ ($r_0$, $\theta _{0}$) + i $\frac{\partial v}{\partial \theta }$($r_0$, $\theta _{0}$)]

after a little of complexed arithmetic which becomes:

    $f' ( z_0)$ =$e^{i\theta_{0} }$/ $r_0$ [$\frac{\partial v}{\partial \theta }$ (r0, $\theta _{0}$) – i $\frac{\partial u}{\partial \theta }$ ($r_0$, $\theta _{0}$)]            (2)

Step III : Both (1) and (2) are giving two expressions for $f'(z_0)$. So when we equate the real and imaginary parts of the right sides of these equations we get the Polar Cauchy-Riemann Equations, that are below:

$\frac{\partial u}{\partial r}$ = $\frac{1}{r}\frac{\partial u}{\partial \theta }$  and

 $\frac{\partial v}{\partial r}$ = -$\frac{1}{r}\frac{\partial u}{\partial \theta }$

Cauchy Riemann Equations Examples

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Problem 1 :    Let $f (z) = y – 2xy + i (-x + x^2 – y^2) + z^2$ with $z = x + iy$. Find the values of ‘$z$’ for which $f'(z)$ exists.

Solution : We have $f (z) = y – 2xy + i ( -x + x^2 – y^2) + z^2$. Also, we are given that $z = x + iy$. Substituting the value of ‘$z$’ in $f (z)$ we get,

$f (z) = y – 2xy + i (-x + x^2 – y^2) + x^2 – y^2 + 2ixy$

= $x^2 – y^2 – 2xy + y + i (-x + x^2 – y^2 + 2xy)$

Let $u (x, y) = x^2 – y^2 – 2xy + y$ and $v (x, y) = -x + x^2 – y^2 + 2xy$

Now compute the partial derivatives of $u$ and $v$, we get,

    $u_x (x, y) = 2x – 2y$,         $v_x (x, y) = -1 + 2x + 2y$

    $u_y (x, y) = -2y – 2x + 1$,    $v_y (x, y) = 2x – 2y$

It is clear that $u_x$ = $v_y$ and $v_x$ = $-u_y$. That is the Cauchy Riemann equations hold true and thus $f' (z)$ exists for every value of ‘z’.

So, $f' (z) = u_x + i v_x = 2x – 2y + i (2x + 2y – 1) = 2z + 2iz – i$

Problem 2 :    Consider the function $g (x + iy) = x^2 – y^2 + 2ixy$ which has $u (x, y) = x^2 – y^2$ and

$v (x, y) = 2xy$. Show that g is an entire function.

Solution: Now evaluating the first order partial derivatives of $u$ and $v$, we get,

$u_x (x, y) = 2x$,             $v_x (x, y) = 2y$

$u_y (x, y) = −2y$,         $v_y (x, y) = 2x$

Clearly, $u_x (x, y)$ = $v_y (x, y)$ and $u_y (x, y) = − v_x (x, y)$. That is the Cauchy Riemann's equations are satisfied for all

$(x, y)$, hence, this function is entire. It is sufficient to observe that

$g (z) = z^2$, since $(x + iy)^2 = x^2 – y^2 + 2ixy$. So g is a polynomial in z and we know that all the polynomials are differentiable at all points.

Problem 3 :    Consider the function $g (x + iy) = i (x^2 + y^2)$ which has $u (x, y)$ = 0 and $v (x, y) = x^2 + y^2$. Using Cauchy

Riemann equations show that function is not analytic.

Solution : So the first order partial derivatives of $u$ and $v$ are:

    $u_x (x, y)$ = 0,        $v_x (x, y) = 2x$

    $u_y (x, y)$ = 0,        $v_y (x, y) = 2y$

We see that the Cauchy Riemann equations, $u_x (x, y)$ = $v_y (x, y)$ and $u_y (x, y) = − v_x (x, y)$ are only satisfied for

points $x$ = 0 and $y$ = 0. So the function is only differentiable at $z$ = 0. So, it not analytic everywhere.

Problem 4 :  For what values Cauchy Riemann's equations are satisfy in $h (z) = e^z$ . Also prove that $h'(z) = e^z$.

Solution: Consider $h (z) = e^z$. So $h (x + iy) = e^x+iy$.

So, $h (z) = e^x (cos y + i sin y)$ = $u (x, y) + i v (x, y)$

So, $u (x, y) = e^x cos y$ and $v (x, y) = e^x sin y$

Evaluating the first order partial derivatives we get,

$u_x (x, y) = e^x cos y$,        $v_x (x, y) = e^x sin y$

$u_y (x, y) = -e^x sin y$,        $v_y (x, y) = e^x cos y$

Clearly, $u_x (x, y)$ = $v_y (x, y)$ and $u_y (x, y)$ = $− v_x (x, y)$. That is the Cauchy Riemann's equations are satisfied for all $(x, y)$, hence, this function is entire.

Thus the derivative of $h (z)$ is:

$h ' (z) = h ' (x + iy)$ 

= $u_x (x, y) + i v_x (x, y)$ 

= $e^x cos y + i e^x sin y$ 

= $e^z$.

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