Cauchy-Riemann Equations

## **Problem**

Consider the complex number in polar form

# z
r

e

i
θ

{\displaystyle z = r{e}^{i \theta} }

, and the complex function in polar form

f
(
z
# )
u
(
r
,
θ
)
i
v
(
r
,
θ
)

{\displaystyle f(z) = u(r, \theta) + iv(r, \theta) }

.

Show that the Cauchy-Riemann equations in polar form are

∂
u

∂
r

=

1
r

∂
v

∂
θ

∂
v

∂
r

=
−

1
r

∂
u

∂
θ

.

{\displaystyle \begin{align}
\frac{\partial u}{\partial r} &= \frac{1}{r} \frac{\partial v}{\partial \theta} \\[5 pt]
\frac{\partial v}{\partial r} &= -\frac{1}{r} \frac{\partial u}{\partial \theta}.
\end{align} }

## **Solution**

It is important to know the Cartesian representation of the complex number

# z
x
i
y

{\displaystyle z = x + iy}

and the Cauchy-Riemann equations

∂
u

∂
x

=

∂
v

∂
y

∂
v

∂
x

=
−

∂
u

∂
y

.

{\displaystyle \begin{align}
\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\[5 pt]
\frac{\partial v}{\partial x} &= -\frac{\partial u}{\partial y}.
\end{align} }

Now use the multivariate chain rule to take the the derivatives of

∂
u

∂
r

{\displaystyle \frac{\partial u}{\partial r} }

,

∂
u

∂
θ

{\displaystyle \frac{\partial u}{\partial \theta} }

,

∂
v

∂
r

{\displaystyle \frac{\partial v}{\partial r} }

, and

∂
v

∂
θ

{\displaystyle \frac{\partial v}{\partial \theta} }

. For the function

u
(
r
,
θ
)

{\displaystyle u(r, \theta)}

:

∂
u

∂
r

=

∂
u

∂
x

∂
x

∂
r

∂
u

∂
y

∂
y

∂
r

{\displaystyle \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} }

∂
u

∂
θ

=

∂
u

∂
x

∂
x

∂
θ

∂
u

∂
y

∂
y

∂
θ

.

{\displaystyle \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}. }

Since

∂
x

∂
r

=
cos
⁡

θ

{\displaystyle \frac{\partial x}{\partial r} = \cos{\theta} }

,

∂
y

∂
r

=
sin
⁡

θ

{\displaystyle \frac{\partial y}{\partial r} = \sin{\theta} }

,

∂
x

∂
θ

=
−
r
sin
⁡

θ

{\displaystyle \frac{\partial x}{\partial \theta} = -r\sin{\theta} }

, and

∂
y

∂
θ

=
r
cos
⁡

θ

{\displaystyle \frac{\partial y}{\partial \theta} = r\cos{\theta} }

, we get

∂
u

∂
r

=

∂
u

∂
x

cos
⁡

θ

∂
u

∂
y

sin
⁡

θ

{\displaystyle \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos{\theta} + \frac{\partial u}{\partial y} \sin{\theta} }

∂
u

∂
θ

=
−

∂
u

∂
x

r
sin
⁡

θ

∂
u

∂
y

r
cos
⁡

θ

.

{\displaystyle \frac{\partial u}{\partial \theta} = -\frac{\partial u}{\partial x} r\sin{\theta} + \frac{\partial u}{\partial y} r\cos{\theta}. }

We repeat the same calculations for

v
(
r
,
θ
)

{\displaystyle v(r, \theta) }

:

∂
v

∂
r

=

∂
v

∂
x

cos
⁡

θ

∂
v

∂
y

sin
⁡

θ

{\displaystyle \frac{\partial v}{\partial r} = \frac{\partial v}{\partial x} \cos{\theta} + \frac{\partial v}{\partial y} \sin{\theta} }

∂
v

∂
θ

=
−

∂
v

∂
x

r
sin
⁡

θ

∂
v

∂
y

r
cos
⁡

θ

.

{\displaystyle \frac{\partial v}{\partial \theta} = -\frac{\partial v}{\partial x} r\sin{\theta} + \frac{\partial v}{\partial y} r\cos{\theta}. }

Now use the Cauchy-Riemann equations in Cartesian form and make the right substitutions.

∂
v

∂
θ

=

∂
u

∂
y

r
sin
⁡

θ

∂
u

∂
x

r
cos
⁡

θ

=
r

∂
u

∂
r

{\displaystyle \frac{\partial v}{\partial \theta} = \frac{\partial u}{\partial y} r\sin{\theta} + \frac{\partial u}{\partial x} r\cos{\theta} = r\frac{\partial u}{\partial r} }

∂
u

∂
θ

=
−

∂
v

∂
y

r
sin
⁡

θ

−

∂
v

∂
x

r
cos
⁡

θ

=
−
r

∂
v

∂
r

{\displaystyle \frac{\partial u}{\partial \theta} = -\frac{\partial v}{\partial y} r\sin{\theta} - \frac{\partial v}{\partial x} r\cos{\theta} = -r\frac{\partial v}{\partial r} }

Therefore,

∂
u

∂
r

=

1
r

∂
v

∂
θ

∂
v

∂
r

=
−

1
r

∂
u

∂
θ

.

{\displaystyle \begin{align}
\frac{\partial u}{\partial r} &= \frac{1}{r} \frac{\partial v}{\partial \theta} \\[5 pt]
\frac{\partial v}{\partial r} &= -\frac{1}{r} \frac{\partial u}{\partial \theta}.
\end{align} }