What if we just.. forgot the i?

The complex numbers $a+bi$ or $u+iv$ are actually not a single number! If you have taken pre calculus or have a nerdy friend who talks about math, you have surely seen the famous complex number written as $a+bi$ .

What is so complex about them?

If you realise $u$ and $v$ are both real numbers, just added with $i = \sqrt{-1}$ , which is infamously called imaginary, then the picture becomes clearer. $u$ steps along our real number line and $v$ steps along the imaginary axis! $v$ itself is real, but it scales $i$ . From here on we will imagine $i$ as a notation to tell us “go perpendicular from the point you are on”.

Can we then study the numbers separately? Yes! $u$ and $v$ are two separate numbers just connected by $i$ . So they must be independent of each other.

…or are they?

Meet $u(x,y)$ and $v(x,y)$ . Yep! We just sneaked in a dependency. Both $u$ and $v$ now depend on $x$ and $y$ . They can still be independent of each other depending on the problem, but both are built from $x$ and $y$ . Imagine them as Bread and Pizza while the independent variables are wheat and water!

So we can now just design it as the following,

f(x,y) = u(x,y) + i \times v(x,y)

Since we took $i$ as just an indicator for perpendicular from a point, you can imagine the points $(x,y,u(x,y))$ and $(x,y,v(x,y))$ are two points in two different 3D planes, where nothing is imaginary anymore!

Yes, after calculating our points we can again put them perpendicularly with the help of our $i$ but the visualisation may be too abstract and not necessary for now.

Stretching is good for health right?

Let us take a non linear $u$ and $v$ . What does it mean? It means for any given $x$ and $y$ , it maps to a number but the change is not linear, it is “curvy”. So if you graph $u$ and $v$ in the complex plane, where $v$ is supposed to be perpendicular to $u$ , you will see it does not even look perpendicular anymore. The curves are just bending around!

So we have this one machine called the Jacobian. It packs all four partial derivatives into one place and tells us how the map is stretching around any given point.

J = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}

But wait, this is a matrix which we study in Linear Algebra right? But $u$ and $v$ are not linear anymore? Yes, if we zoom in closely to a point on the graph we can see them slowly turn linear, and that is how the Jacobian lets us study the nature of these curves.

Here is a wonderful proof of how the Jacobian Matrix works by Mathemaniac on YT

Our honest try to find the derivative of $f(x,y) = u(x,y) + iv(x,y)$ gives us this matrix which does tells us about how it behaves around the point but there are 4 numbers to tell us one fact! Is it differentiable or not??

The problem nobody tells you

Taking a derivative must be ONE number. But the machine gives us FOUR! What if we wanted to differentiate $f$ as a complex function, not as two real ones?

f'(z) = \lim_{h \rightarrow 0} \frac{f(z+h) - f(z)}{h}

The complex plane is pretty weird so the limit above is not very intuitive. But mathematicians have a favourite trick: fix one direction at a time. Think of it as walking on the $X$ number line only and walking towards the point. Then do the same on the $Y$ number line only. In both the ways you will realise you have reached the same place in the same way, IF the derivative existed. More formally,

\begin{align*} &\text{let } h = \Delta x : f'(z) = \frac{\partial u}{\partial x} + i \times \frac{\partial v}{\partial x} \\ &\text{let } h = i\Delta y : f'(z) = \left[\frac{\partial u}{\partial y} + i \times \frac{\partial v}{\partial y}\right] / i = \frac{\partial v}{\partial y} - i \times \frac{\partial u}{\partial y} \end{align*}

Visual by complex-analysis.com

And we just learnt if the complex derivative exists, however you go they tend to be the same which means,

\begin{align*} & &\frac{\partial u}{\partial x} + i \times \frac{\partial v}{\partial x} &= \frac{\partial v}{\partial y} - i \times \frac{\partial u}{\partial y} \\ &\implies &u_x + iv_x &= v_y + i(-u_y) \\ &\implies &u_x = v_y&, v_x = -u_y \end{align*}

What we realise with the last two equations is that, we cannot just throw random $u$ and $v$ into $f$ and ask it to be a smooth differentiable map! Instead ONLY the $u$ and $v$ which satisfies the last 2 equations or as mathematician loves to say “constraints” are valid

Survivors

So those two constraints are not just gatekeepers. The functions that survive them have something very cool going on. Let us plug the constraints back into the Jacobian and see what falls out.

The Jacobian gave us a way to see how our map stretches, and the derivative method gave us two constraints. Putting those values in,

J = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix} = \begin{pmatrix} u_x & -v_x\\ v_x & u_x \end{pmatrix}

Ah, the matrix looks too ugly, let us give it nicknames $A = u_x$ and $B = v_x$ , then

J = \begin{pmatrix} a & -b\\ b & a \end{pmatrix}

Have you ever seen this matrix before? Let us remember how we said $i$ means perpendicular, right? What if we multiply a number with a complex number?

Okay, let us remember how we said having an $i$ can mean that the number is perpendicular, right? What if we multiply a number with a complex number?

Take $a + i \times b = r_1 e^{i\theta_1}$ and $x + i \times y = r_2 e^{i\theta_2}$ . We take the Euler form because unlike junky numbers, Euler form talks with us in angles.

\begin{align*} (a+ib)(x+iy) &= (ax−by) + i(bx+ay) \\ &= r_{1} r_{2} e^{i(\theta_1 + \theta_2)} \end{align*}

The Euler form tells us the new product has a scaled size of $r_1 \times r_2$ and a new rotation of $\theta_1 + \theta_2$ . So multiplying two complex numbers just scales and rotates into a new number, no shear.

Now look at what we have, $(ax-by)$ and $(bx+ay)$ . Let us make that silly matrix out of them:

\begin{pmatrix} a & -b\\ b & a \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} ax - by\\ bx + ay \end{pmatrix}

We DID NOT intentionally design our Jacobian this way, but we arrived at the understanding that any $u(x,y)$ and $v(x,y)$ we choose, if we want $f(x,y)$ to be derivable, must obey this. So the Jacobian of any surviving function is a transformation which just scales and rotates input values $(x,y)$ with absolutely no shear. Only then we can proudly say our function is derivable in Complex!

Also, this is about the right time to mention our MVP. The constraints we found are famously known as the \verb|Cauchy-Riemann equations| which are necessary conditions to call a complex function differentiable.