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Analyticity of a function

Short answer

Writing sin z = sin x cosh y + i cos x sinh y, the partial derivatives are continuous and satisfy u_x=v_y, u_y=-v_x everywhere. Hence sin z is entire with derivative cos z.

For a complex variable, show that f(z) = sin z is analytic.

A function f(z) = u + iv is analytic in a region if u and v have continuous first partial derivatives there and satisfy the Cauchy–Riemann equations u_x = v_y and u_y = −v_x.

Write z = x + iy and use sin(x + iy) = sin x cos(iy) + cos x sin(iy) = sin x cosh y + i cos x sinh y.

u = sin x cosh y,    v = cos x sinh y

u_x = cos x cosh y    v_y = cos x cosh y   → u_x = v_y ✓
u_y = sin x sinh y    v_x = −sin x sinh y  → u_y = −v_x ✓

The partial derivatives are continuous for all x, y, and the Cauchy–Riemann equations hold everywhere. Hence sin z is analytic everywhere (it is an entire function), and f'(z) = u_x + i v_x = cos x cosh y − i sin x sinh y = cos z.

✓ Answer checked by Claude Opus 5.5 (high effort)

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