What are Wirtinger derivatives?

Wirtinger derivatives are differential operators defined for a complex variable z = x + iy as d/dz = (1/2)(d/dx - i d/dy) and d/dz* = (1/2)(d/dx + i d/dy). They allow differentiating functions of complex variables directly without decomposing into real and imaginary parts, and are standard tools in complex optimization for signal processing and machine learning.

In what contexts are Wirtinger derivatives used?

They are used in deriving the Wiener filter (the optimal solution satisfies d epsilon/dG* = 0), backpropagation in complex-valued neural networks, update rules for adaptive filters (LMS/RLS), and beamforming in MIMO communications. Their greatest advantage is that the optimality condition for a real-valued function f(z, z*) can be written concisely as df/dz* = 0.

Wirtinger Derivatives

Introduction

Wirtinger derivatives are differential operators for functions depending on complex variables, regardless of whether those functions are real-valued or complex-valued. Named after Wilhelm Wirtinger, this formalism treats functions on the complex plane as functions of two formally independent variables, $z$ and $z^*$. It is particularly useful in optimization problems involving non-holomorphic functions (especially real-valued functions), enabling complex optimization problems to be handled in the same way as their real counterparts.

Why are Wirtinger derivatives needed?

Ordinary complex differentiation (for holomorphic functions satisfying the Cauchy--Riemann equations) cannot handle non-holomorphic functions such as the real-valued function $f(z) = |z|^2$. Yet in signal processing and machine learning, one frequently needs to minimize real-valued loss functions that depend on complex parameters. Wirtinger derivatives provide the natural framework for such problems.

Although the formalism can be applied to complex-valued functions as well, for holomorphic functions ordinary complex differentiation suffices. The true power of Wirtinger derivatives emerges when dealing with functions that depend on both $z$ and $z^*$ (i.e., non-holomorphic functions).

Definition

Consider a function $f(z, z^*)$ depending on the complex variable $z = x + iy$ ($x, y \in \bbR$), where $z^* = x - iy$ denotes the complex conjugate of $z$.

Definition (Wirtinger differential operators)

For $z = x + iy$, the Wirtinger differential operators are defined as follows:

\begin{align} \dfrac{\partial}{\partial z} &= \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} - i\dfrac{\partial}{\partial y}\right) \label{eq:wirtinger-z}\\ \dfrac{\partial}{\partial z^*} &= \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\right) \label{eq:wirtinger-zbar} \end{align}

Meaning of the factor $\dfrac{1}{2}$

The factor $\dfrac{1}{2}$ is chosen so that $\dfrac{\partial z}{\partial z} = 1$. Indeed, since $z = x + iy$:

$$\dfrac{\partial z}{\partial z} = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} - i\dfrac{\partial}{\partial y}\right)(x + iy) = \dfrac{1}{2}(1 + 1) = 1$$

Expression in terms of real and imaginary part derivatives

Solving equations (\ref{eq:wirtinger-z}) and (\ref{eq:wirtinger-zbar}) simultaneously, one can express the real and imaginary part derivatives in terms of Wirtinger derivatives:

\begin{align} \dfrac{\partial}{\partial x} &= \dfrac{\partial}{\partial z} + \dfrac{\partial}{\partial z^*}\\ \dfrac{\partial}{\partial y} &= i\left(\dfrac{\partial}{\partial z} - \dfrac{\partial}{\partial z^*}\right) \end{align}

Basic Properties

Theorem (basic differentiation formulas)

For complex scalars $z$ and $w$:

\begin{align} \dfrac{\partial z}{\partial z} &= 1, & \dfrac{\partial z}{\partial z^*} &= 0 \label{eq:basic-diff-1}\\ \dfrac{\partial z^*}{\partial z} &= 0, & \dfrac{\partial z^*}{\partial z^*} &= 1 \label{eq:basic-diff-2}\\ \dfrac{\partial |z|^2}{\partial z} &= z^*, & \dfrac{\partial |z|^2}{\partial z^*} &= z \label{eq:abs-z-sq-diff}\\ \dfrac{\partial (z^* w)}{\partial z} &= 0, & \dfrac{\partial (z^* w)}{\partial z^*} &= w \label{eq:product-diff} \end{align}

Proof example: $\dfrac{\partial}{\partial z^*} |z|^2 = z$

Step 1: Express $|z|^2$ in terms of real and imaginary parts.

Setting $z = x + iy$, we have $z^* = x - iy$. Therefore:

\begin{equation} |z|^2 = z z^* = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 - i^2 y^2 = x^2 + y^2 \label{eq:proof-step1} \end{equation}

Step 2: Apply the definition of the Wirtinger differential operator to $|z|^2$.

Applying the Wirtinger operator from equation \eqref{eq:wirtinger-zbar} to $|z|^2 = x^2 + y^2$ from equation \eqref{eq:proof-step1}:

\begin{equation} \dfrac{\partial}{\partial z^*} |z|^2 = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\right) (x^2 + y^2) \label{eq:proof-step2} \end{equation}

Here, the left-hand side denotes the operator $\dfrac{\partial}{\partial z^*}$ acting on the function $|z|^2$.

Step 3: Compute the partial derivative with respect to $x$.

Differentiating $(x^2 + y^2)$ in equation \eqref{eq:proof-step2} with respect to $x$:

\begin{equation} \dfrac{\partial}{\partial x}(x^2 + y^2) = 2x + 0 = 2x \label{eq:proof-step3} \end{equation}

Step 4: Compute the partial derivative with respect to $y$.

Differentiating $(x^2 + y^2)$ in equation \eqref{eq:proof-step2} with respect to $y$:

\begin{equation} \dfrac{\partial}{\partial y}(x^2 + y^2) = 0 + 2y = 2y \label{eq:proof-step4} \end{equation}

Step 5: Substitute the results.

Substituting equations \eqref{eq:proof-step3} and \eqref{eq:proof-step4} into equation \eqref{eq:proof-step2}:

\begin{align} \dfrac{\partial}{\partial z^*} |z|^2 &= \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\right) (x^2 + y^2) \tag{\ref{eq:proof-step2}}\\ &= \dfrac{1}{2}\left[\dfrac{\partial}{\partial x}(x^2 + y^2) + i\dfrac{\partial}{\partial y}(x^2 + y^2)\right] \tag{distributing the operator}\\ &= \dfrac{1}{\cancel{2}}\left[\cancel{2}x + i \cdot \cancel{2}y\right] \tag{substituting \ref{eq:proof-step3}, \ref{eq:proof-step4}}\\ &= x + iy \label{eq:proof-step5-4}\\ &= z \label{eq:proof-step5-5} \end{align}

Thus, $\dfrac{\partial}{\partial z^*} |z|^2 = z$ has been established. $\square$

Relationship with the Cauchy--Riemann equations

Theorem (characterization of holomorphicity)

A function $f(z)$ is holomorphic (satisfies the Cauchy--Riemann equations) if and only if:

\begin{equation} \dfrac{\partial f}{\partial z^*} = 0 \label{eq:holomorphic-condition} \end{equation}

Why does "independent of $z^*$" imply holomorphicity?

To understand this equivalence, one must properly grasp the Wirtinger derivative framework.

1. The Wirtinger derivative framework

In the Wirtinger framework, $z$ and $z^*$ are treated as formally independent variables. Although this may seem strange at first (since $z^*$ is the complex conjugate of $z$ and hence not independent), this framework allows any function $f(z, z^*)$ to be expressed as a function of two variables $z$ and $z^*$.

Examples:

$|z|^2 = z z^*$ depends on both $z$ and $z^*$
$z^2 + z$ depends only on $z$ (not on $z^*$)
$e^z \sin z$ depends only on $z$

2. Three equivalent definitions of a holomorphic function

(a) Complex differentiability

$f(z)$ is holomorphic at $z_0$ $\Leftrightarrow$ the complex derivative $\displaystyle \lim_{h \to 0} \dfrac{f(z_0+h) - f(z_0)}{h}$ exists

Here $h \in \bbC$, and the limit must be the same regardless of the direction from which $h$ approaches $0$ in the complex plane. This is a far stronger condition than differentiability for real functions (where only limits from the left and right are required).

(b) Cauchy--Riemann equations

$f(z) = u(x,y) + iv(x,y)$ is holomorphic $\Leftrightarrow$ the following holds:

\begin{equation} \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}, \quad \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x} \label{eq:cauchy-riemann} \end{equation}

(c) Characterization via Wirtinger derivatives

$f(z)$ is holomorphic $\Leftrightarrow$ the following holds:

$$\dfrac{\partial f}{\partial z^*} = 0 \tag{\ref{eq:holomorphic-condition}}$$

3. Why (b) and (c) are equivalent

Using the definition of the Wirtinger operator:

$$\dfrac{\partial}{\partial z^*} = \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\right) \tag{\ref{eq:wirtinger-zbar}}$$

and applying it to $f = u + iv$:

\begin{align} \dfrac{\partial f}{\partial z^*} &= \dfrac{1}{2}\left(\dfrac{\partial}{\partial x} + i\dfrac{\partial}{\partial y}\right)(u + iv) \label{eq:holomorphic-proof-1}\\ &= \dfrac{1}{2}\left[\left(\underbrace{\dfrac{\partial u}{\partial x} - \dfrac{\partial v}{\partial y}}_{=0}\right) + i\left(\underbrace{\dfrac{\partial v}{\partial x} + \dfrac{\partial u}{\partial y}}_{=0}\right)\right] \label{eq:holomorphic-proof-2}\\ &= 0 \end{align}

By equation \eqref{eq:holomorphic-condition}, this expression vanishes if and only if:

Real part $= 0$: $\dfrac{\partial u}{\partial x} - \dfrac{\partial v}{\partial y} = 0$ $\Rightarrow$ $\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$
Imaginary part $= 0$: $\dfrac{\partial v}{\partial x} + \dfrac{\partial u}{\partial y} = 0$ $\Rightarrow$ $\dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}$

These are precisely the Cauchy--Riemann equations from equation \eqref{eq:cauchy-riemann}!

4. Intuitive understanding

In the framework where $z$ and $z^*$ are treated as independent variables:

A general function can be written as $f(z, z^*)$ (e.g., $|z|^2 = zz^*$)
A holomorphic function can be expressed in terms of $z$ alone (e.g., $z^2$, $e^z$, $\sin z$)
"Independent of $z^*$" = "unchanged by $\dfrac{\partial}{\partial z^*}$" = equation \eqref{eq:holomorphic-condition}

Therefore, equation \eqref{eq:holomorphic-condition} states that "$f$ is a function of $z$ alone," which is the essence of holomorphicity.

Optimization of Real-Valued Functions

The most important application of Wirtinger derivatives is the optimization of real-valued functions that depend on complex parameters.

Theorem (stationary point condition)

Let $f: \bbC \to \bbR$ be a real-valued function. The following three conditions are equivalent:

$$z_0 \text{ is a stationary point of } f \quad \Leftrightarrow \quad \dfrac{\partial f}{\partial z^*}\bigg|_{z=z_0} = 0 \quad \Leftrightarrow \quad \dfrac{\partial f}{\partial z}\bigg|_{z=z_0} = 0$$

Proof

Step 1: Definition of a stationary point

A stationary point $z_0$ of $f: \mathbb{C} \to \mathbb{R}$ is a point where the gradient vanishes:

\begin{equation} \dfrac{\partial f}{\partial x}\bigg|_{z=z_0} = 0 \text{ and } \dfrac{\partial f}{\partial y}\bigg|_{z=z_0} = 0 \label{eq:stationary-def} \end{equation}

Since $f$ is real-valued, both $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ are real.

Step 2: Equivalence with $\dfrac{\partial f}{\partial z^*} = 0$

By equation \eqref{eq:wirtinger-zbar}:

\begin{equation} \dfrac{\partial f}{\partial z^*} = \dfrac{1}{2}\left(\dfrac{\partial f}{\partial x} + i\dfrac{\partial f}{\partial y}\right) \label{eq:wirtinger-zbar-app} \end{equation}

Since $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ are real, equation \eqref{eq:wirtinger-zbar-app} equals $0$ if and only if both the real and imaginary parts vanish, which is equivalent to equation \eqref{eq:stationary-def}.

Step 3: Equivalence with $\dfrac{\partial f}{\partial z} = 0$

By equation \eqref{eq:wirtinger-z}:

\begin{equation} \dfrac{\partial f}{\partial z} = \dfrac{1}{2}\left(\dfrac{\partial f}{\partial x} - i\dfrac{\partial f}{\partial y}\right) \label{eq:wirtinger-z-app} \end{equation}

Similarly, since $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ are real, equation \eqref{eq:wirtinger-z-app} equals $0$ if and only if equation \eqref{eq:stationary-def} holds.

Conclusion

By Steps 2 and 3, both $\dfrac{\partial f}{\partial z^*} = 0$ and $\dfrac{\partial f}{\partial z} = 0$ are equivalent to the stationary condition \eqref{eq:stationary-def}. $\square$

Important identity

When $f$ is a real-valued function, the following relationship holds between the Wirtinger derivatives:

\begin{equation} \dfrac{\partial f}{\partial z} = \left( \dfrac{\partial f}{\partial z^*} \right)^* \label{eq:wirtinger-conjugate} \end{equation}

Therefore, if $\dfrac{\partial f}{\partial z^*} = 0$, then $\dfrac{\partial f}{\partial z} = 0$ follows automatically (since the complex conjugate of $0$ is $0$). In practice, it is conventional to solve $\dfrac{\partial f}{\partial z^*} = 0$.

Optimization example

Example (complex least squares problem)

Given complex numbers $a, b \in \mathbb{C}$ ($a \neq 0$), find the value of $z$ that minimizes $f(z) = |az - b|^2$.

Solution:

Step 1: Expand $f$ as a polynomial in $z$ and $z^*$

Using $|w|^2 = w w^*$ with $w = az - b$:

\begin{align} f(z, z^*) &= (az - b)(az - b)^* \label{eq:ex-step1}\\ &= (az - b)(a^* z^* - b^*) \label{eq:ex-step2} \end{align}

Expanding:

\begin{equation} f(z, z^*) = \underbrace{a a^*}_{|a|^2} z z^* - a b^* z - a^* b z^* + \underbrace{b b^*}_{|b|^2} \label{eq:ex-expanded} \end{equation}

Step 2: Compute the Wirtinger derivative

Differentiating each term of equation \eqref{eq:ex-expanded} with respect to $z^*$. By equation \eqref{eq:abs-z-sq-diff}, $\dfrac{\partial}{\partial z^*}(zz^*) = z$, so:

\begin{align} \dfrac{\partial f}{\partial z^*} &= |a|^2 \underbrace{\dfrac{\partial}{\partial z^*}(z z^*)}_{=z} - a b^* \underbrace{\dfrac{\partial z}{\partial z^*}}_{=0} - a^* b \underbrace{\dfrac{\partial z^*}{\partial z^*}}_{=1} + 0 \label{eq:ex-diff1}\\ &= |a|^2 z - a^* b \label{eq:ex-diff2} \end{align}

Step 3: Solve the stationary condition

Setting $\dfrac{\partial f}{\partial z^*} = 0$:

\begin{align} |a|^2 z &= a^* b \nonumber\\ z &= \dfrac{a^* b}{|a|^2} \label{eq:ex-solution} \end{align}

Extension to Vectors and Matrices

Wirtinger derivatives extend naturally to vectors and matrices.

Definition (Wirtinger derivative for vectors)

For $\bm{z} = [z_1, \ldots, z_n]^T \in \bbC^n$ and a real-valued function $f(\bm{z}, \bm{z}^*)$:

$$\dfrac{\partial f}{\partial \bm{z}^*} = \begin{bmatrix} \dfrac{\partial f}{\partial z_1^*} \\ \vdots \\ \dfrac{\partial f}{\partial z_n^*} \end{bmatrix} \in \bbC^n$$

Useful formulas

Theorem (vector and matrix differentiation formulas)

For $\bm{z}, \bm{w} \in \bbC^n$, $\bm{A} \in \bbC^{n \times n}$ (Hermitian matrix), and $\bm{b} \in \bbC^n$:

\begin{align} \dfrac{\partial}{\partial \bm{z}^*}(\bm{w}^H \bm{z}) &= \bm{0} \label{eq:vec-wHz}\\ \dfrac{\partial}{\partial \bm{z}^*}(\bm{z}^H \bm{w}) &= \bm{w} \label{eq:vec-zHw}\\ \dfrac{\partial}{\partial \bm{z}^*}(\bm{z}^H \bm{A} \bm{z}) &= \bm{A} \bm{z} \label{eq:vec-zHAz}\\ \dfrac{\partial}{\partial \bm{z}^*}(\bm{z}^H \bm{b} + \bm{z}^H \bm{A} \bm{z}) &= \bm{b} + \bm{A} \bm{z} \label{eq:vec-combined} \end{align}

Here, $\bm{w}^H$ denotes the conjugate transpose of $\bm{w}$. For proofs, see Proofs in Matrix Calculus, 16.57.

Example (complex Wiener--Hopf equation)

Minimize the following quadratic form over the complex vector $\bm{z} \in \bbC^n$:

$$J(\bm{z}) = \bm{z}^H \bm{R} \bm{z} - \bm{z}^H \bm{p} - \bm{p}^H \bm{z} + \sigma^2$$

where $\bm{R} = \bm{R}^H$ (Hermitian matrix), $\bm{p} \in \bbC^n$, and $\sigma^2 \in \bbR$.

Solution:

Step 1: Differentiate each term with respect to $\bm{z}^*$

$J$ consists of four terms. Using equations \eqref{eq:vec-wHz}--\eqref{eq:vec-zHAz}, we differentiate each one.

\begin{align} \dfrac{\partial}{\partial \bm{z}^*}(\bm{z}^H \bm{R} \bm{z}) &= \bm{R} \bm{z} & &\text{(Eq.}\;\eqref{eq:vec-zHAz}\text{: }\bm{A} = \bm{R}\text{)} \label{eq:wh-term1}\\ \dfrac{\partial}{\partial \bm{z}^*}(-\bm{z}^H \bm{p}) &= -\bm{p} & &\text{(Eq.}\;\eqref{eq:vec-zHw}\text{: }\bm{w} = \bm{p}\text{)} \label{eq:wh-term2}\\ \dfrac{\partial}{\partial \bm{z}^*}(-\bm{p}^H \bm{z}) &= \bm{0} & &\text{(Eq.}\;\eqref{eq:vec-wHz}\text{: }\bm{w} = \bm{p}\text{)} \label{eq:wh-term3}\\ \dfrac{\partial}{\partial \bm{z}^*}(\sigma^2) &= \bm{0} & &\text{(constant)} \label{eq:wh-term4} \end{align}

Step 2: Sum the results

Adding equations \eqref{eq:wh-term1}--\eqref{eq:wh-term4}:

\begin{equation} \dfrac{\partial J}{\partial \bm{z}^*} = \bm{R} \bm{z} - \bm{p} + \bm{0} + \bm{0} = \bm{R} \bm{z} - \bm{p} \label{eq:wh-gradient} \end{equation}

Step 3: Solve the stationary condition

Setting $\dfrac{\partial J}{\partial \bm{z}^*} = \bm{0}$:

\begin{align} \bm{R} \bm{z} &= \bm{p} \nonumber\\ \bm{z}_{\text{opt}} &= \bm{R}^{-1} \bm{p} \label{eq:wh-solution} \end{align}

This is the solution of the complex Wiener filter (Wiener--Hopf equation). Since $\bm{R}$ is a Hermitian positive definite matrix, its inverse exists.

Worked Examples

Example 1: Stationary points of $f(z) = |z|^4$

Find the stationary points of $f(z, z^*) = (z z^*)^2$.

Solution:

Step 1: Express $f$ in terms of $z$ and $z^*$

Since $|z|^2 = z z^*$:

\begin{equation} f(z, z^*) = (z z^*)^2 = |z|^4 \label{eq:ex1-step1} \end{equation}

Step 2: Differentiate with respect to $z^*$

Using the chain rule:

\begin{align} \dfrac{\partial f}{\partial z^*} &= \dfrac{\partial}{\partial z^*}(z z^*)^2 \nonumber\\ &= 2(z z^*) \cdot \dfrac{\partial}{\partial z^*}(z z^*) & &\text{(chain rule)} \label{eq:ex1-step2a}\\ &= 2(z z^*) \cdot z & &\text{(Eq.}\;\eqref{eq:abs-z-sq-diff}\text{)} \label{eq:ex1-step2b}\\ &= 2|z|^2 z \label{eq:ex1-step2c} \end{align}

Step 3: Solve the stationary condition

Setting $\dfrac{\partial f}{\partial z^*} = 0$:

\begin{equation} 2|z|^2 z = 0 \label{eq:ex1-step3} \end{equation}

Since $|z|^2 \geq 0$, this equation holds only when $z = 0$.

Conclusion

$z = 0$ is the unique stationary point. Since $f(z) = |z|^4 \geq 0$ and $f(0) = 0$, it is the global minimum.

Example 2: Minimizing $f(z) = |z - a|^2$

Minimize the squared distance $f(z) = |z - a|^2$ between a point $z$ in the complex plane and a fixed point $a \in \bbC$.

Solution:

Step 1: Express $f$ in terms of $z$ and $z^*$

Using $|w|^2 = w w^*$:

\begin{align} f(z, z^*) &= |z - a|^2 = (z - a)(z - a)^* \nonumber\\ &= (z - a)(z^* - a^*) \nonumber\\ &= z z^* - z a^* - a z^* + a a^* \label{eq:ex2-step1} \end{align}

Step 2: Differentiate with respect to $z^*$

Differentiating each term of equation \eqref{eq:ex2-step1} with respect to $z^*$:

\begin{align} \dfrac{\partial f}{\partial z^*} &= \dfrac{\partial}{\partial z^*}(z z^* - z a^* - a z^* + a a^*) \nonumber\\ &= z - 0 - a + 0 & &\text{(applying Eq.}\;\eqref{eq:abs-z-sq-diff}\text{ to each term)} \label{eq:ex2-step2a}\\ &= z - a \label{eq:ex2-step2b} \end{align}

Key point: Differentiating with respect to $z^*$, terms containing $z$ yield $z$, while terms not containing $z^*$ (such as $z a^*$) are treated as constants.

Step 3: Solve the stationary condition

Setting $\dfrac{\partial f}{\partial z^*} = 0$:

\begin{equation} z - a = 0 \quad \Rightarrow \quad z = a \label{eq:ex2-step3} \end{equation}

Step 4: Verify minimality via the Hessian matrix

Writing $z = x + iy$, we compute the Hessian matrix in the real variables $(x, y)$. Since $f = (x - a_x)^2 + (y - a_y)^2$:

\begin{equation} H = \begin{pmatrix} \dfrac{\partial^2 f}{\partial x^2} & \dfrac{\partial^2 f}{\partial x \partial y} \\ \dfrac{\partial^2 f}{\partial y \partial x} & \dfrac{\partial^2 f}{\partial y^2} \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = 2I \label{eq:ex2-hessian} \end{equation}

Since $H$ is positive definite (all eigenvalues are positive), $z = a$ is a strictly convex minimum.

Conclusion

$z = a$ is the unique stationary point and the global minimum (with minimum value $f(a) = 0$).

Geometric interpretation: The point in the complex plane closest to a fixed point $a$ is $a$ itself -- an intuitively obvious result that is derived algebraically via Wirtinger derivatives.

Example 3: Two-variable case

Minimize $f(z_1, z_2) = |z_1 - z_2|^2 + |z_1|^2$.

Solution:

Step 1: Expand $f$

Using $|w|^2 = w w^*$:

\begin{align} f &= (z_1 - z_2)(z_1 - z_2)^* + z_1 z_1^* \nonumber\\ &= (z_1 - z_2)(z_1^* - z_2^*) + z_1 z_1^* \nonumber\\ &= z_1 z_1^* - z_1 z_2^* - z_2 z_1^* + z_2 z_2^* + z_1 z_1^* \label{eq:ex3-step1} \end{align}

Step 2: Differentiate with respect to $z_1^*$

Differentiating each term of equation \eqref{eq:ex3-step1} with respect to $z_1^*$:

\begin{align} \dfrac{\partial f}{\partial z_1^*} &= z_1 - z_2 - 0 + 0 + z_1 \nonumber\\ &= 2z_1 - z_2 \label{eq:ex3-step2} \end{align}

Step 3: Differentiate with respect to $z_2^*$

Differentiating each term of equation \eqref{eq:ex3-step1} with respect to $z_2^*$:

\begin{align} \dfrac{\partial f}{\partial z_2^*} &= 0 - 0 - z_1 + z_2 + 0 \nonumber\\ &= z_2 - z_1 \label{eq:ex3-step3} \end{align}

Step 4: Solve the stationary conditions

Setting $\dfrac{\partial f}{\partial z_1^*} = 0$ and $\dfrac{\partial f}{\partial z_2^*} = 0$:

\begin{align} 2z_1 - z_2 &= 0 \label{eq:ex3-step4a}\\ z_2 - z_1 &= 0 \label{eq:ex3-step4b} \end{align}

From equation \eqref{eq:ex3-step4b}, $z_2 = z_1$. Substituting into equation \eqref{eq:ex3-step4a}:

$$2z_1 - z_1 = z_1 = 0$$

Conclusion

$z_1 = z_2 = 0$ is the unique stationary point. Since $f(0, 0) = 0$ and $f(z_1, z_2) \geq 0$, it is the global minimum.

Applications

Signal Processing

Adaptive filters: Gradient computation in the complex LMS algorithm
Wiener filter: Derivation of the optimal linear estimator for complex signals
Array signal processing: Beamforming optimization

Machine Learning

Complex-valued neural networks: Gradient descent with respect to complex parameters
Quantum machine learning: Optimization of quantum states with complex amplitudes

Communication Theory

MIMO communications: Optimal receiver design for complex channel matrices
Equalizers: Optimal detection for complex symbols