Suppose that $f(z) = \frac{z}{1-z}$. Use Cauchy-Riemann Equations to determine the analyticity of $f(z)$. Then find the derivative of $f(z)$.

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can anyone help me in solving this question? I have tried expressing $z=x+iy$ and multiplying the conjugate to numerator and denominator. However, the expansion and differentiation are too tedious. Is there an easier way to solve this?

Answer 1

$$f(z) = \frac{z}{1-z} = \frac{x+iy}{1-x-iy} \cdot \frac{1-x+iy}{1-x+iy} = \frac{x-x^2-y^2 + iy}{(1-x)^2 + y^2}$$ Let $$u(x,y) = \frac{x-x^2-y^2}{(1-x)^2 + y^2}$$ and $$v(x,y) = \frac{y}{(1-x)^2 + y^2}$$ Then, $f = u + iv$. I have done the messy part of the calculation for you. I leave it to you to check whether or not the Cauchy-Riemann equations hold, and comment on the differentiability of $u,v: \mathbb R^2\to\mathbb R$. These are straightforward computations since $u(x,y)$ and $v(x,y)$ are known. Note that $f$ is defined only for $z\ne 1$, so $x = 1$ and $y = 0$ are not permitted.

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Edit $1$: You can compute the partial derivatives using the quotient rule. You can verify your calculations with the following: $$u_x = \frac{x^2- 2x -y^2 + 1}{((1-x)^2 + y^2)^2}$$ $$u_y = \frac{2(x-1)y}{((1-x)^2 + y^2)^2}$$ $$v_x = \frac{2(1-x)y}{((1-x)^2 + y^2)^2}$$ $$v_y = \frac{x^2- 2x -y^2 + 1}{((1-x)^2 + y^2)^2}$$