1. **Check analyticity of** $f(z) = e^{z^2}$ using Cauchy-Riemann equations. 2. Write $z = x + iy$, then $f(z) = u(x,y) + iv(x,y)$ where $$f(z) = e^{(x+iy)^2} = e^{x^2 - y^2 + 2ixy} = e^{x^2 - y^2}(\cos(2xy) + i\sin(2xy))$$ So, $$u = e^{x^2 - y^2} \cos(2xy), \quad v = e^{x^2 - y^2} \sin(2xy)$$ 3. Compute partial derivatives: $$u_x = e^{x^2 - y^2}(2x \cos(2xy) - 2y \sin(2xy))$$ $$u_y = e^{x^2 - y^2}(-2y \cos(2xy) - 2x \sin(2xy))$$ $$v_x = e^{x^2 - y^2}(2x \sin(2xy) + 2y \cos(2xy))$$ $$v_y = e^{x^2 - y^2}(-2y \sin(2xy) + 2x \cos(2xy))$$ 4. Check Cauchy-Riemann equations: $$u_x = v_y \implies 2x \cos(2xy) - 2y \sin(2xy) = -2y \sin(2xy) + 2x \cos(2xy)$$ $$u_y = -v_x \implies -2y \cos(2xy) - 2x \sin(2xy) = - (2x \sin(2xy) + 2y \cos(2xy))$$ Both hold for all $x,y$. 5. Since Cauchy-Riemann equations hold everywhere and $u,v$ have continuous partial derivatives, $f$ is analytic everywhere. 6. The complex derivative is: $$f'(z) = \frac{d}{dz} e^{z^2} = 2z e^{z^2}$$ **Final answer:** $f$ is analytic everywhere and $f'(z) = 2z e^{z^2}$.