Calculus 3... Using the Cauchy-Riemann equations, show that ∂v/∂r=-1/r ∂u/∂θ

Let x=rcosθ and y=rsinθ. Suppose that u and v are functions u=f(x,y) and v=g(x,y) that satisfy the Cauchy-Riemann Equations ∂u/∂x=∂v/∂y , ∂u/∂y=-∂v/∂x.

Show that ∂v/∂r= -1/r ∂u/∂θ

I'm struggling with this one and any help would be greatly appreciated, Thanks!!

∂v/∂r = ∂v/∂x ∂x/∂r + ∂v/∂y ∂y/∂r, by the Chain Rule
.......= ∂v/∂x * (cos θ) + ∂v/∂y * (sin θ)
.......= -∂u/∂y * (cos θ) + ∂u/∂x (sin θ), by Cauchy-Riemann Equations

Therefore,
∂u/∂θ = ∂u/∂x ∂x/∂θ + ∂u/∂y ∂y/∂θ, by the Chain Rule
.......= ∂u/∂x * (-r sin θ) + ∂u/∂y * (r cos θ)
.......= -r [∂u/∂x * (sin θ) - ∂u/∂y * (cos θ)]
.......= -r ∂v/∂r, by first computation.

Hence, ∂v/∂r = (-1/r) ∂u/∂θ.

I hope this helps!

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