THE CAUCHY-RIEMANN EQUATION SAMUEL ALEXANDRE VIDAL CYRUS DIVECHA

1. Foreword Today I want to expand on a remark one of us made about holomorphic functions. What are holomorphic functions, why are they so stunning, what makes them so useful ? Definition 1.1. A holomorphic function f (z) is a complex valued function of the complex variable such that f (z + h) − f (z) h has a limit when h tends to zero without being equal to zero. Of course if the limit exists then its value is denoted by f 0 (z) and is called the derivative of f at z. You may think this is simply the definition of differentiability we all learned when we were children, but here the variable h is supposed to be complex and the limit has to exist and be the same independently of the way h tends to zero.

2. Differentiability in the real variable sense Consider a complex valued function f (z) of the complex variable z, defined in an open set U . This means that the function f associates to any complex number z in the open set U a complex number f (z). A first question is : what is the possible meaning of d f (z) ? First let’s rephrase that in the real variable sense. We recall that complex numbers are expressions like z = x + iy where x and y are both real numbers and i is a square root of −1, which means simply that we compute with the extra relation i2 = −1. Such a function can be written f (x + i y) = f1 (x, y) + i f2 (x, y) where f1 and f2 are two real valued functions of the real variables x and y. The function f1 (x, y) is the real part of f and f2 (x, y) is its imaginary part. 1

Example 1. f (z) = z 3 = (x + i y)3 = x3 + 3 iyx2 − 3 xy 2 − i y 3 = x3 − 3 xy 2 + i (3 yx2 − y 3 ) = f1 + i f2 where f1 (x, y) = x3 − 3 xy 2 and f2 (x, y) = 3 yx2 − y 3 where f1 (x, y) the real part and f2 (x, y) is the imaginary part of f (z) = z 3 . Following the book we can write,     ∂f1 ∂f1 ∂f2 ∂f2 d f (x, y) = (x, y) dx + (x, y) dy + i (x, y) dx + (x, y) dy ∂x ∂y ∂x ∂y There are still complex number concepts in this formula. The first step in the ‘reification’ of our function f was to consider it a function of the vector with real coordinates x and y. The second and last step in this reification is to consider it vector valued.     f1 (x, y) x = f f2 (x, y) y So now d f can be written in matrix form with only real numbers.   ∂f1 ∂f1 (x, y) (x, y)    ∂x   ∂y x   dx df = (1)  y  ∂f2  dy ∂f2 (x, y) (x, y) ∂x ∂y I know some of you may ask what the hell are those d, dx, dy in the formula. A complete answer would be a bit lengthy but genuinely interesting. Let me tell you a story : when I still was at the University, there were barely no one to answer that question with the right answer while almost everyone had its own interpretation. Some were close to be true, some were dangerously wrong. We may have another conversation at the caf´e on that topic. The expressions dx and dy are differential forms (this is some kind of mathematical objects, they may look like functions but it’s dangerously wrong to think that way). You can take them as new extra variables and compute with them as if they were some coordinates. The thing that has to be remembered is the pullback formula (2)

d f (x) = f 0 (x) dx

By the way, this is the same dx you see in the integrals. Let’s move on. Here dx and dy are the coordinates of a tangent vector at the point of coordinate (x, y) and d f (x, y) is the corresponding tangent vector when 2

transformed through f . With formula 1 you can see one important thing : the coordinates of d f (x, y) linearly depend on the coordinates (dx, dy). Hence the matrix in the middle. That matrix is called the jacobian matrix (after the name of Jacobi). So that’s it ? Is it what it means to be holomorphic ? To have partial derivatives with respect to the real coordinates so that the jacobian matrix is defined ?   ∂f1 ∂f1 (x, y) (x, y)  ∂x  ∂y   df =    ∂f2  ∂f2 (x, y) (x, y) ∂x ∂y Actually no. All this means, if those partial derivatives are continuous, is that the function f is differentiable in the real variable sense. Example 2. Consider the function f (z) = z¯

3. Differentiability in the comlex variable sense To get a good notion of differentiability in the complex variable sense, one key point is the pullback formula (formula 2 in the previous section). More concretely, one wants that the derivative of a complex valued function of the complex variable is itself a complex function of the complex variable. This way, f 0 (z) has to be a complex number, which means that the jacobian matrix has to transform the complex plane as it were multiplication by a complex number. Recall that any complex number z = x + i y behave as a real matrix. From the product (x + i y)(x0 + i y 0 ) = (xx0 − yy 0 ) + i (xy 0 + yx0 ) one can see the linear relation on real coordinates :  0     xx − yy 0 x −y x0 = xy 0 + yx0 y x y0 Again you can see that this matrix can’t be any matrix, there are constraints on its entries. The two coefficient of the main diagonal are equal and the two coefficients of the other diagonal are opposed. If you want the jacobian matrix of the previous section to have the exact same structure it means that you have a system of two relations.  ∂f1 ∂f2   (x, y) = (x, y)  ∂x ∂y ∂f ∂f    2 (x, y) = − 1 (x, y) ∂x ∂y Those simple relations are called the Cauchy-Riemann equations. Theorem 3.1. Let f be a complex function of the complex variable, then the two following conditions are equivalent, i.

the function f is holomorphic. 3

ii.

the function f is real differentiable and satisfies the Cauchy-Riemann equations.

Exercise 3.1. Those who cares are invited to prove the equivalence, personally I don’t. Let’s consider now some examples and counter examples. Example 3. What is the jacobian matrix of the function f (z) = z 3 ? With z = x + i y as before, ∂f1 ∂f1 (x, y) = 3 x2 − 3 y 2 , (x, y) = −6 xy, ∂x ∂y ∂f2 ∂f2 (x, y) = 6 xy, (x, y) = 3 x2 − 3 y 2 . ∂x ∂y So the jacobian matrix is,   2 3 x − 3 y2 −6 xy df = 6 xy 3 x2 − 3 y 2 Here we are, the Cauchy-Riemann equations are satisfied and the jacobian matrix looks like the multiplication by f 0 (x + i y) = (3 x2 − 3 y 2 ) + i 6 xy. Just expand the expression 3 z 2 = 3 (x + i y)2 and see this is equal to f 0 (z). Example 4. Now consider the function f (z) = z¯ that is, f (x + i y) = x − i y. the jacobian matrix is,   1 0 df = 0 −1 The Cauchy-Riemann equations are not satisfied, so according to Theorem 3.1, the limit of Definition 1.1 cannot exist. And indeed, if you look at the limit of, f (z + h) − f (z) h when h tends to zero with values only on the real axis (without zero) you find 1, while the limit of the same expression when h tends to zero with values only on the imaginary axis (without zero) you find −1. Those two quantities are different and so the function can’t be holomorphic.

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