# Complex derivative as a real derivative?

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?

• What is the $2 \times 1$ vector given by $f'(z)$? – Gibbs Dec 10 '17 at 21:09
• What do you mean by "don't match up"? – Ted Shifrin Dec 10 '17 at 21:09
• @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? – ManishKumar Singh Dec 10 '17 at 21:12
• @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 – ManishKumar Singh Dec 10 '17 at 21:13
• @ManishKumarSingh so if $f(z) = z^2$ what do you get? – Gibbs Dec 10 '17 at 21:15

## 2 Answers

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.

• @TedShifrin I thought about it, and I concluded that I stand by what I said originally. – Arthur Dec 10 '17 at 21:24
• @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. – ManishKumar Singh Dec 10 '17 at 21:37
• @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. – Arthur Dec 10 '17 at 21:46
• @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. – ManishKumar Singh Dec 10 '17 at 21:50
• @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. – 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.