[proofplan] Multiplication by $f'(w) \neq 0$ rotates every tangent vector by the same angle $\arg f'(w)$. The angle between two curves at $w$ is the difference of their tangent arguments, and the [Chain Rule](/theorems/323) gives each image tangent as $f'(w) \cdot \gamma_i'(0)$. The common rotation factor cancels in the difference of arguments, preserving the angle. [/proofplan] [step:Compute the tangent vectors of the image curves via the chain rule] The angle of the tangent vector of $\gamma_i$ at $w$ is $\arg \gamma_i'(0)$, and the angle between the two curves at $w$ is \begin{align*} \theta = \arg \gamma_2'(0) - \arg \gamma_1'(0). \end{align*} The image curves $\tilde{\gamma}_i = f \circ \gamma_i$ satisfy, by the [Chain Rule](/theorems/323): \begin{align*} \tilde{\gamma}_i'(0) = f'(\gamma_i(0)) \cdot \gamma_i'(0) = f'(w) \cdot \gamma_i'(0). \end{align*} Since $f'(w) \neq 0$ and $\gamma_i'(0) \neq 0$, we have $\tilde{\gamma}_i'(0) \neq 0$, so the image curves have well-defined tangent directions. [/step] [step:Show the rotation factor $\arg f'(w)$ cancels in the angle difference] The angle between the image curves at $f(w)$ is: \begin{align*} \tilde{\theta} &= \arg \tilde{\gamma}_2'(0) - \arg \tilde{\gamma}_1'(0) \\ &= \arg\bigl(f'(w) \cdot \gamma_2'(0)\bigr) - \arg\bigl(f'(w) \cdot \gamma_1'(0)\bigr) \\ &= \bigl(\arg f'(w) + \arg \gamma_2'(0)\bigr) - \bigl(\arg f'(w) + \arg \gamma_1'(0)\bigr) \\ &= \arg \gamma_2'(0) - \arg \gamma_1'(0) = \theta. \end{align*} The key identity is $\arg(ab) = \arg a + \arg b$ (modulo $2\pi$) for nonzero complex numbers $a, b$. The common factor $\arg f'(w)$ cancels, so the angle -- and its orientation (sign) -- are preserved. [guided] Why does complex [differentiability](/page/Derivative) with $f'(w) \neq 0$ imply conformality, while real differentiability does not? The [Cauchy--Riemann Characterisation](/theorems/333) shows that the Jacobian of a holomorphic function has the form $\begin{pmatrix} \alpha & -\beta \\ \beta & \alpha \end{pmatrix}$, which represents multiplication by the complex number $\alpha + i\beta = f'(w)$. This matrix factors as $|f'(w)| \cdot R_\theta$ where $\theta = \arg f'(w)$ and $R_\theta$ is a rotation matrix. A rotation-plus-scaling preserves angles between vectors. A general real-differentiable map has an arbitrary $2 \times 2$ Jacobian, which can shear and stretch differently in different directions, distorting angles. The Cauchy--Riemann equations $u_x = v_y$, $u_y = -v_x$ constrain the Jacobian to the special conformal form. The only [linear maps](/page/Linear%20Map) $\mathbb{R}^2 \to \mathbb{R}^2$ that correspond to complex multiplication are rotations-with-scaling (or zero). The computation in the proof is direct: $\arg(f'(w) \cdot \gamma_i'(0)) = \arg f'(w) + \arg \gamma_i'(0)$ by the multiplicativity of $\arg$ for nonzero complex numbers. Taking the difference of two such expressions, the common factor $\arg f'(w)$ cancels, leaving the original angle $\arg \gamma_2'(0) - \arg \gamma_1'(0)$. The condition $f'(w) \neq 0$ is essential: if $f'(w) = 0$, the image tangent vectors $\tilde{\gamma}_i'(0) = f'(w) \cdot \gamma_i'(0) = 0$, and no angle is defined. At a zero of $f'$ of order $k$, the function typically multiplies angles by $k + 1$. [/guided] [/step]