I'm looking for the maximum value of the modulus of a holomorphic function, and I am getting a bit stuck.
The function is $$(z-1)\left(z+\frac{1}{2}\right)$$ with domain $\,|z| \leq 1\,$
Now, I know by the maximum modulus principle the max value will occur on the boundary. So by multiplying the two expressions I get:
$$\left|z^2 – \frac{1}{2}z – \frac{1}{2}\right|$$
writing in complex polar form (and applying MMP, so $\,r = 1\,$) I then get:
$$\left|e^{2i\theta} -\frac{1}{2}\,e^{i\theta}-\frac{1}{2}\right|$$
And… this is where I am stuck. So any help would be greatly appreciated!
Best Answer
$$f(z) = (z-1)(z+1/2) = z^2 - z/2 - 1/2$$ Since you know that the maximum is hit on the boundary, $z = e^{i\theta}$, we get that $$F(\theta) = e^{2i \theta} - \dfrac{e^{i\theta}}2 - \dfrac12 = \left(\cos(2\theta) - \dfrac{\cos(\theta)}2 - \dfrac12 \right)+i \left(\sin(2 \theta) - \dfrac{\sin(\theta)}2\right)$$ Let $g(\theta) = \vert F(\theta) \vert^2$. \begin{align} g(\theta) & = \vert F(\theta) \vert^2 = \left(\cos(2\theta) - \dfrac{\cos(\theta)}2 - \dfrac12 \right)^2 + \left(\sin(2 \theta) - \dfrac{\sin(\theta)}2\right)^2 \\ & = \cos^2(2\theta) + \dfrac{\cos^2(\theta)}4 + \dfrac14 - \cos(2\theta) \cos(\theta) - \cos(2\theta) + \dfrac{\cos(\theta)}2\\ & + \sin^2(2\theta) + \dfrac{\sin^2(\theta)}4 - \sin(2\theta) \sin(\theta)\\ & = 1 + \dfrac14 + \dfrac14 - \dfrac{\cos(\theta)}2 - \cos(2\theta)\\ & = \dfrac32 - \dfrac{\cos(\theta)}2 - \cos(2 \theta)\\ & = \dfrac52 - \dfrac{\cos(\theta)}2 - 2 \cos^2(\theta)\\ & = \dfrac52 - 2 \left( \cos(\theta) + \dfrac18\right)^2 + 2 \left(\dfrac18 \right)^2\\ & = \dfrac52 + \dfrac1{32} - 2 \left( \cos(\theta) + \dfrac18\right)^2 \end{align} The maximum is at $\cos(\theta) = -\dfrac18$ and the maximum value of $g(\theta) = \dfrac{81}{32}$. Hence, the maximum value is $$\max_{\vert z \vert \leq 1}\vert f(z) \vert = \sqrt{\dfrac{81}{32}} = \dfrac98 \sqrt{2}$$