11.2: Tangent vectors as complex numbers

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In previous classes, you used parametrized curves \(\gamma (t) = (x(t), y(t))\) in the \(xy\)-plane. Considered this way, the tangent vector is just the derivative:

\[\gamma '(t) = (x' (t), y' (t)). \nonumber \]

Note, as a vector, \((x', y')\) represents a displacement. If the vector starts at the origin, then the endpoint is at \((x', y')\). More typically we draw the vector starting at the point \(\gamma (t)\).

You may also previously used parametrized curves \(\gamma (t) = x(t) + iy(t)\) in the complex plane. Considered this way, the tangent vector is just the derivative:

\[\gamma '(t) = x' (t) + iy' (t). \nonumber \]

It should be clear that these representations are equivalent. The vector \((x', y')\) and the complex number \(x' + iy'\) both represent the same displacement. Also, the length of a vector and the angle between two vectors is the same in both representations.

Thinking of tangent vectors to curves as complex numbers allows us to recast conformality in terms of complex numbers.

Theorem \(\PageIndex{1}\)

If \(f(z)\) is conformal at \(z_0\) then there is a complex number \(c = ae^{i \phi}\) such that the map \(f\) multiplies tangent vectors at \(z_0\) by \(c\). Conversely, if the map \(f\) multiplies all tangent vectors at \(z_0\) by \(c = ae^{i \phi}\) then \(f\) is conformal at \(z_0\).

Proof: By definition \(f\) is conformal at \(z_0\) means that there is an angle \(\phi\) and a scalar \(a > 0\) such that the map \(f\) rotates tangent vectors at \(z_0\) by \(\phi\) and scales them by \(a\). This is exactly the effect of multiplication by \(c = ae^{i \phi}\).