Question

Given f(z) = x²+y²-2x+4+i(2xy-2y)
Express both f(z) and f'(z) in terms of the complex variable z=x+iy

Answer 1

Answer on Question #79195 – Math – Complex Analysis

Question

Given $f(z) = x^2 + y^2 - 2x + 4 + i(2xy - 2y)$

Express both $f(z)$ and $f'(z)$ in terms of the complex variable $z = x + iy$

Solution

By condition

$$u(x, y) = \operatorname{Re} f(z) = x^2 + y^2 - 2x + 4$$ $$v(x, y) = \operatorname{Im} f(z) = 2xy - 2y$$

Then

$$\frac{\partial u}{\partial x} = 2x - 2$$ $$\frac{\partial^2 u}{\partial x^2} = 2$$ $$\frac{\partial u}{\partial y} = 2y$$ $$\frac{\partial^2 u}{\partial y^2} = 2$$

Then

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 4 \neq 0$$

But then function $u(x, y)$ is not harmonious and there is no function $f(z)$ with the real part $x^2 + y^2 - 2x + 4$.

Answer: there is no such function.

(perhaps there is a mistake in the condition and the function has the form $x^2 - y^2 - 2x + 4 + i(2xy - 2y)$, then we can find $f(z)$).

For example,

$$f(z) = x^2 - y^2 - 2x + 4 + i(2xy - 2y)$$

Then

$$u(x, y) = \operatorname{Re} f(z) = x^2 - y^2 - 2x + 4$$ $$v(x, y) = \operatorname{Im} f(z) = 2xy - 2y$$ $$\frac{\partial u}{\partial x} = 2x - 2$$ $$\frac{\partial v}{\partial x} = 2y$$ $$f(z) = f(x + iy) = x^2 - y^2 - 2x + 4 + i(2xy - 2y) = x^2 + 2ixy - y^2 - 2x - 2iy + 4 = (x + iy)^2 - 2(x + iy) + 4 = z^2 - 2z + 4$$ $$f'(z) = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = 2x - 2 + i \cdot 2y = 2(x + iy) - 2 = 2z - 2$$

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