Part (a) Suppose $f'(z)$ is a complex derivation of $f(z)$. Since $f'(z)$ takes complex values, $f'(z)$ is a $2 \times 1$ column vector in $\mathbb{R}^2$.

Part (b) If I interpret $f$ as function between $\mathbb{R}^2 \rightarrow \mathbb{R}^2$, then the derivation $f'(z)$ becomes a $2 \times 2$ linear tranformation.

Conclusion of Part (a) and Part (b) don't match up. Why?

  • $\begingroup$ What is the $2 \times 1$ vector given by $f'(z)$? $\endgroup$ – Gibbs Dec 10 '17 at 21:09
  • $\begingroup$ What do you mean by "don't match up"? $\endgroup$ – Ted Shifrin Dec 10 '17 at 21:09
  • $\begingroup$ @TedShifrin: f'(x) is 2 x 1 matrix in part (a) and 2 x 2 matrix in part (b). How do you explain this inconsistency? $\endgroup$ – ManishKumar Singh Dec 10 '17 at 21:12
  • $\begingroup$ @Gibbs It depends on what f(z) is. but we know for sure f'(z) is a complex function, so it should be of the form 2 x 1 matirx $\endgroup$ – ManishKumar Singh Dec 10 '17 at 21:13
  • $\begingroup$ @ManishKumarSingh so if $f(z) = z^2$ what do you get? $\endgroup$ – Gibbs Dec 10 '17 at 21:15

The real derivative of a holomorphic function (which is required for the complex derivative to exist) at a point is the product of some multiple of the identity matrix with a rotation matrix (this is the reason for the name "holomorphic": it preserves angles, unless the derivative is $0$). Thus only a $2$-dimensional subspace of the $2\times2$ matrices appear. Specifically, only matrices of the form $$ \begin{bmatrix}a&-b\\b&a\end{bmatrix} $$ for some real $a,b$ appear in this setting. Alternatively, the Cauchy-Riemann equations are satisfied iff the real derivative has exactly this form.

The corresponding complex derivative is $f'(z)=a+bi$, or $$\begin{bmatrix}a\\b\end{bmatrix}$$in your matrix form.

  • $\begingroup$ @TedShifrin I thought about it, and I concluded that I stand by what I said originally. $\endgroup$ – Arthur Dec 10 '17 at 21:24
  • $\begingroup$ @Arthur: If I understood you correctly, you are saying 2x2 matrix is of rank 1 and hence can be reduced to 2x1 matrice. If that's true, then should both the map induce same linear transf. $\endgroup$ – ManishKumar Singh Dec 10 '17 at 21:37
  • $\begingroup$ @ManishKumarSingh The $2\times 2$ matrix above has determinant $a^2+b^2$, and is therefore either invertible or $0$, so $1$ is the only rank it can't have. Also, they do induce the same linear map. Remember that the $2\times1$ matrix is really a complex number, so you can multiply two of them together, and the result is the same as if you let the left one be the $2\times2$ matrix and use regular matrix multiplication. This way they are both points in the plane and linear maps of the plane at the same time. $\endgroup$ – Arthur Dec 10 '17 at 21:46
  • $\begingroup$ @Arthur: So what's happening here is we are making use of the fact that $R^2$ is field in complex situation where in real-situation its only a vector space. $\endgroup$ – ManishKumar Singh Dec 10 '17 at 21:50
  • $\begingroup$ @ManishKumarSingh That, and the fact that holomorphic functions are very limited so that the space of possible derivatives is two-dimensional. You could even say that making this work is how we define complex multiplication. $\endgroup$ – Arthur Dec 10 '17 at 22:00

They don't match up because they are different objects. This should be familiar: for functions $f:\mathbb R\to \mathbb R,$ $f'(x)$ is a number, while $Df(x)$ is the unique linear transformation from $\mathbb R$ to $\mathbb R$ such that $f(x+h)-f(x) = Df(x)(h) + o(h).$ We do have $Df(x)(h) = f'(x)h$ for all $h,$ so the two objects have a direct connection, but they shouldn't be confused with each other, and therefore the same notation, $f'(x),$ shouldn't be used for both of them - at least not until the difference between the two is fully absorbed.

Moving to $\mathbb C= \mathbb R^2,$ things will again be confusing if the notation $f'(z)$ is used for two different objects, namely the complex derivative, which is the number $f'(z),$ and the corresponding (real) linear transformation $Df(z):\mathbb R^2\to \mathbb R^2.$ It's the same "problem" as in the first paragraph. There's no actual problem unless we get careless with the notation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.