Question

Prove that sin z is one-to-one on the domain defined by the inequalities

Answer 1

Answer:

By definition we have

sin(z)=12ı⋅(eız−e−ız)

Since the sum of two analytic functions is analytic, it suffices to show that z↦eız and z↦e−ız are analytic. Let z:=x+ıy (x,y∈R), then

eız=eıx⋅e−y=e−y⋅cos(x)=:u(x,y)+ıe−y⋅sinx=:v(x,y)

From

∂xu(x,y)=−e−y⋅cos(x)=∂yv(x,y)∂yu(x,y)=−e−y⋅cos(x)=−∂xv(x,y)

we see that the Cauchy-Riemann equations are satisfied. Since the partial derivatives exist (and are continuous) we conclude that z↦eız is analytic. A similar argumentation shows that z↦e−ız is analytic.