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Comparing the difference quotient along the real and the imaginary direction gives $u_x = v_y$ and $u_y = -v_x$: necessary for differentiability everywhere, sufficient on an open set with continuous partials, and equivalent to the statement that $f$ does not depend on the conjugate.
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 derive the Cauchy-Riemann equations by comparing two directions, compute the four partial derivatives of a given function and test them, state what the equations do and do not guarantee in each direction, find the derivative from the partials, and identify the open set on which a function is analytic — including the functions for which that set is empty.
The complex derivative as a limit that must agree along every approach, and the observation that a function needing a conjugate to write down seems never to be differentiable. This lesson compares just two of those approaches and gets two equations, and those two equations turn the observation into a test.
For $f = u + iv$, the four partial derivatives $u_x, u_y, v_x, v_y$ are taken with the other variable held fixed. The Cauchy-Riemann equations are $u_x = v_y$ and $u_y = -v_x$. A set is open when every point of it has a disc about it inside the set, and connected when any two of its points are joined by a path inside it; a region or domain is an open connected set.
Where the equations come from. Suppose $f'(z_0)$ exists. Compute the limit twice, with $h$ real and with $h$ purely imaginary:
$$f'(z_0) = u_x + iv_x \qquad\text{and}\qquad f'(z_0) = v_y - iu_y.$$
They must agree, so comparing real and imaginary parts gives
$$u_x = v_y, \qquad u_y = -v_x.$$
These are necessary: wherever a complex derivative exists, they hold.
They are not sufficient on their own. Two directions agreeing never proved a limit, and it does not prove one here. The repair is a theorem: if the four partial derivatives exist, are continuous on an open set, and satisfy the two equations there, then $f$ is analytic on that set, with
$$f' = u_x + iv_x = v_y - iu_y.$$
In practice the partials are continuous whenever anyone is looking, so the test is check the two equations on an open set.
What the equations say geometrically. The real derivative of $f$ as a map of the plane is the matrix $\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}$, and the equations say exactly that this matrix has the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ — which is multiplication by the complex number $a + ib$. Among all linear maps of the plane, those are precisely the rotations combined with a scaling. So complex differentiability is the demand that the local linear approximation be a rotation and a stretch rather than a general linear map.
And in terms of the conjugate. Writing $x$ and $y$ in terms of $z$ and $\bar z$, the pair is equivalent to the single equation $\partial f / \partial \bar z = 0$: $f$ does not depend on $\bar z$. That is the rule of thumb of the last lesson, exactly stated.
Another way: picture
Draw a tiny square at $z_0$ and look at its image. A general smooth map sends it to a small parallelogram, which may be sheared and may be flipped over. The Cauchy-Riemann equations say the image is a square again — turned and resized, never sheared and never flipped. Every later theorem about angles being preserved is this picture, and so is the fact that conjugation, which flips, is analytic nowhere.
Another way: steps
The useful form of the question is never is this function analytic but on what set is it analytic, and the four cases that come up are worth having in mind as a list.
Everywhere. Polynomials in $z$, the exponential, sine and cosine. Both equations hold identically, and the function is entire.
Everywhere except some isolated points. Rational functions, away from the zeros of the denominator. This is the case unit 4 is about: the exceptional points are the singularities, and almost all the information in the function turns out to be stored at them.
On a cut plane. A branch of the logarithm, or of a fractional power. The equations hold wherever the branch is continuous, which is everywhere except the cut.
On a set with no interior — so nowhere. $|z|^{2}$, whose equations hold at the origin alone. $\bar z$ and $\operatorname{Re} z$, whose equations hold at no point. Also $f(z) = \bar z^{2}$, and every other function built with a conjugate.
That last case is worth dwelling on, because it is the one where the word analytic does real work. Differentiability at a point is a fact about one point and buys nothing. Every theorem from here to the end of the course asks for an open set, and a function differentiable at a single point satisfies none of them. 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.
The implication proved in this lesson runs one way: differentiable at a point implies the equations hold there. The converse is false, and the standard counterexample is worth carrying.
Let $f(z) = \sqrt{|xy|}$, real-valued. At the origin all four partial derivatives are zero, so both equations hold. But approach along the line $y = x$ and the difference quotient does not settle: $f$ is not differentiable at the origin. Two directions agreed and told us nothing, exactly as in the limits lesson.
This is why the sufficient version of the theorem carries two extra hypotheses — an open set and continuous partial derivatives — and why a solution that checks the equations at one point and announces therefore differentiable has skipped the part of the argument that is actually hard.
The related slip is to check the equations, find they hold on a line or at a point, and call the function analytic there. Analytic means differentiable throughout a disc. A line has no disc inside it.
$f(z) = e^{z} = e^{x}\cos y + ie^{x}\sin y$, so $u = e^{x}\cos y$ and $v = e^{x}\sin y$.
Split into the two parts.
$u_x = e^{x}\cos y$ and $v_y = e^{x}\cos y$: equal. $u_y = -e^{x}\sin y$ and $v_x = e^{x}\sin y$: opposite.
Both equations, at every point.
The partials are continuous, so $f$ is entire, and $f' = u_x + iv_x = e^{x}\cos y + ie^{x}\sin y = e^{z}$.
The derivative comes out of the same partials.
$f(z) = |z|^{2}$: $u = x^{2} + y^{2}$, $v = 0$.
Split into the two parts.
$u_x = 2x$ must equal $v_y = 0$, and $u_y = 2y$ must equal $-v_x = 0$. Both hold only at the origin.
One point.
A single point contains no disc, so there is no open set on which $f$ is differentiable: analytic nowhere.
Analytic is a word about a neighbourhood.
$u_x = 2x$ and $v_y = 2y$, so the first equation says $x = y$; $u_y = 0$ and $v_x = 0$, so the second holds everywhere.
One equation is free, the other is a condition.
So it is differentiable exactly on the line $y = x$, which contains no disc: analytic nowhere.
For $f(z) = z^{2}$ we have $u = x^{2} - y^{2}$ and $v = 2xy$. Fill in the table at the point $x = 5$, $y = 8$.
| Value | |
|---|---|
| The real part of the function at the point | |
| The imaginary part of the function at the point | |
| The partial of the real part in x | |
| The partial of the real part in y | |
| The partial of the imaginary part in x | |
| The partial of the imaginary part in y |
For $f = u + iv$, is this the Cauchy-Riemann pair: $u_{xx} + u_{yy} = 0$ on its own?
$f(z) = z^{3}$, so $u = x^{3} - 3xy^{2}$. What is $u_x$ at $x = 3$, $y = 6$?
Answer:
Select every statement that is true.
This task has no paper form; do it on a device.
Match each function to the set of points at which it has a derivative.
| Every point of the plane | No point at all | The origin and nowhere else | Every point except the origin | |
|---|---|---|---|---|
| $z^{7}$ | ||||
| $\bar z$ | ||||
| $|z|^{2}$ | ||||
| $\dfrac{1}{z}$ |
$f$ and $\bar f$ are both analytic on a disc. How many different values can $f$ take on that disc?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(z) = z^{2}$ we have $u = x^{2} - y^{2}$ and $v = 2xy$. Fill in the table at the point $x = 7$, $y = 8$.
| Value | |
|---|---|
| The real part of the function at the point | |
| The imaginary part of the function at the point | |
| The partial of the real part in x | |
| The partial of the real part in y | |
| The partial of the imaginary part in x | |
| The partial of the imaginary part in y |
You can test a function for analyticity with the two equations and say what extra hypotheses the converse needs. Say in your own words why satisfying the equations at a single point is not enough. Next: harmonic functions, which are what the equations force the two parts to be.
9. Your turn: where is $f(z) = x^{2} + iy^{2}$ differentiable?, step 2