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The usual difference quotient with a complex increment, which keeps every differentiation rule and changes what the rules are worth: the conjugate, the real part and the squared modulus have no derivative anywhere, because the limit has to survive every direction of approach.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the definition of the complex derivative, use the power, sum, product, quotient and chain rules, show from the definition that the conjugate is differentiable nowhere, distinguish differentiability at a point from analyticity on an open set, and say why complex differentiability is a far stronger demand than differentiability as a map of the plane.
Limits in the plane, and the fact that a limit has to agree along every approach. A derivative is a limit of that kind, so everything that made limits demanding makes derivatives far more demanding still — and that is the whole reason this subject looks nothing like real calculus after the first few pages.
$f$ is differentiable at $z_0$ when the difference quotient has a limit there, written $f'(z_0)$. $f$ is analytic (equivalently holomorphic) on an open set when it is differentiable at every point of it, and analytic at a point when it is analytic on some disc about that point — a stronger statement than being differentiable at the point alone. A function analytic on the whole plane is entire.
The definition.
$$f'(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h},$$
with $h$ a complex increment tending to $0$ in the plane. The formula is the one from real calculus; the quantifier behind it is not.
Everything algebraic survives. Sum, product, quotient and chain rules hold, with the proofs copied out unchanged, because those proofs used only the algebra of limits. So $(z^{n})' = nz^{n-1}$, $(e^{z})' = e^{z}$, $(\sin z)' = \cos z$, and a branch of $\log$ has derivative $1/z$. Every polynomial is entire; every rational function is analytic off its poles.
And differentiability still implies continuity, by the same three-line argument: the difference of the values is the quotient times the increment, and a limit times zero is zero.
What does not survive is how easy it is to be differentiable. Consider $f(z) = \bar z$. Its difference quotient is $\bar h / h$, which is $1$ for real $h$ and $-1$ for imaginary $h$. So the conjugate — a map that is smooth, distance-preserving and invertible when read as a map of $\mathbb{R}^{2}$ — has a derivative at no point whatever.
The same happens to $\operatorname{Re} z$, to $\operatorname{Im} z$ and to $|z|^{2}$ away from the origin. The rule of thumb, and it is exactly right: a function built from $z$ alone is differentiable; a function that needs $\bar z$ to write down is not. The next lesson turns that rule of thumb into a pair of equations.
Analytic is a word about a neighbourhood. $|z|^{2}$ is differentiable at the origin and at no other point, so it is analytic nowhere. Differentiability at a single point buys almost nothing; differentiability on an open set buys the rest of this course.
Another way: picture
Fix $z_0$ and let $h$ run round a tiny circle about the origin. The difference quotient traces some little curve, and differentiability says that curve collapses to a single point as the circle shrinks. For a polynomial it does. For the conjugate the quotient $\bar h / h$ runs round the whole unit circle for every radius, however small: shrinking $h$ does nothing at all. That picture is what the two-directions computation is sampling.
Another way: steps
A real function can be differentiable once and not twice: $x|x|$ is the standard example. A real function can be infinitely differentiable and not equal to its own Taylor series anywhere but one point. Neither can happen here.
A function analytic on an open set is automatically infinitely differentiable on it, and automatically equal to its Taylor series on every disc inside it. Its values on any small circle determine it everywhere inside. If it is bounded on the whole plane it is constant. If two analytic functions agree on a segment, they agree everywhere they are both defined.
None of that is proved until unit 3, and none of it is plausible from the definition. But it is worth knowing now, because it explains why so much effort goes into deciding whether a function is analytic: the word is not a technical precondition, it is the entire hypothesis of every theorem to come. Complex differentiability is not the real kind with a letter added. The difference quotient has to settle down to the same number along every one of the infinitely many directions a point can be approached from in a plane, and that single demand is strong enough to force a function to have derivatives of every order, to equal its own Taylor series, and to be determined on a whole region by its values on a curve. Nothing in real calculus behaves like that, so a fact carried over from it without checking is a guess.
Read $f = u + iv$ as a map from $\mathbb{R}^{2}$ to $\mathbb{R}^{2}$ and ask whether it is differentiable in the multivariable sense: whether it is well approximated near each point by a linear map. Conjugation passes that test easily — it is a linear map, the reflection in the real axis.
Complex differentiability asks for more. It asks that the approximating linear map be multiplication by a complex number, which among all linear maps of the plane are exactly the rotations-with-scaling. Reflection is not one of them: it reverses orientation, and no rotation does.
So the question is never is this map nice but is the derivative a complex number. That is the content of the Cauchy-Riemann equations in the next lesson, and it is why a course in complex analysis is not a course in the calculus of two real variables with different notation.
One smaller habit to watch: differentiable at $z_0$ and analytic at $z_0$ are not the same claim, and almost every theorem in this course needs the second. Saying analytic when only one point has been checked is the most common way to get a true-sounding argument that proves nothing.
$f(z) = z^{2}$: the difference quotient is $\dfrac{(z_0 + h)^{2} - z_0^{2}}{h} = \dfrac{2z_0h + h^{2}}{h} = 2z_0 + h$.
Expand and cancel the increment.
As $h \to 0$ this tends to $2z_0$ — and it does so along every approach, because $h$ itself is what is becoming small.
No direction was privileged.
$f(z) = |z|^{2} = z\bar z$. The quotient at $z_0$ is $\dfrac{(z_0 + h)\overline{(z_0 + h)} - z_0\bar z_0}{h} = \bar z_0 + \bar h + z_0\dfrac{\bar h}{h}$.
Expand with the conjugate.
The first two terms behave; the last is $z_0$ times something with no limit. So unless $z_0 = 0$ there is no derivative.
The troublesome term has a coefficient.
At $z_0 = 0$ the quotient is just $\bar h \to 0$. So $f$ is differentiable at the origin, with derivative $0$, and nowhere else — hence analytic nowhere.
One point is not an open set.
The quotient is $\dfrac{\operatorname{Re}(z_0 + h) - \operatorname{Re} z_0}{h} = \dfrac{\operatorname{Re} h}{h}$.
The point drops out again.
Real $h$ gives $1$; imaginary $h$ gives $0$. No limit, at any point.
Let $f(z) = 7z^{6}$. Fill in the table.
| Value | |
|---|---|
| The coefficient of the first derivative | |
| The exponent in the first derivative | |
| The coefficient of the second derivative | |
| The exponent in the second derivative | |
| The coefficient of the third derivative | |
| The exponent in the third derivative |
$f(z) = z^{2} + z + 2$. What is $f'(6)$?
Answer:
Put the steps of showing that $f(z) = \bar z$ has a derivative at no point in order.
Number the steps in order (write the number in the box):
Match each function to its derivative.
| $4z^{3}$ | $7e^{7z}$ | $-\dfrac{1}{z^{2}}$ | $\dfrac{1}{z}$ | |
|---|---|---|---|---|
| $z^{4}$ | ||||
| $e^{7z}$ | ||||
| $\dfrac{1}{z}$ | ||||
| a branch of $\log z$ |
$f(z) = z^{8}$ and $|z| = 1$. What is $|f'(z)|$?
Answer:
Build the proof that a function differentiable at a point is continuous there.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $f(z) = 3z^{7}$. Fill in the table.
| Value | |
|---|---|
| The coefficient of the first derivative | |
| The exponent in the first derivative | |
| The coefficient of the second derivative | |
| The exponent in the second derivative | |
| The coefficient of the third derivative | |
| The exponent in the third derivative |
You can differentiate any function written in $z$ alone, and you can show from the definition that a function written with a conjugate has no derivative. Say in your own words what the difference quotient is being asked to do. Next: the Cauchy-Riemann equations, which turn that demand into two equations you can check.
9. Your turn: is $f(z) = \operatorname{Re} z$ differentiable anywhere?, step 2