Find the maximum points of $|z^2-3z+2|$ at $|z|\leq 1$
Because $z^2-3z+2$ the function is analytic on the interior of $|z|\leq 1$ the maximum will be obtained on the boundary.
Let $z=e^{it}$ where $0\leq t \leq 2\pi$
So the function is $e^{2it}-3e^{it}+2$ now I want to look at the boundry so I should take $|e^{2it}|-3|e^{it}|+2$?
On
If you mean the maximum of $\lvert z^2-3z+2\rvert=\lvert z-1\rvert\lvert z-2\rvert$, on the closed unit disc, a simple sketch will convince you the maximum of each factor is attained at $z=-1$, so the maximum of their product is attained at the same point, and it's equal to $6$.
On
Hint: If you mean the maximum magnitude of that expression, then you want to find $t$ maximizing $$f(t)=|e^{2it}-3e^{it}+2|^2$$ $$=|(\cos 2t-3\cos t+2)+i(\sin 2t-3\sin t)|^2$$ $$=(\cos 2t-3\cos t+2)^2 + (\sin 2t-3\sin t)^2$$ (note that it is enough to maximize the square of the magnitude; that's convenient for getting rid of the pesky square root).
Addendum: It's easy to differentiate $f$ and factor the result to get $$f'(t) = 8(\sin t)(\cos t - \tfrac32)(\cos ^2 t - \tfrac32)$$
Since both of the last two factors are strictly negative for all $t$, this vanishes precisely when $\sin t$ vanishes, i.e., when $t=k\pi$ for integral $k$. You can check that $$f(k\pi)=\begin{cases} =0, & k \textrm{ even} & \textrm{(so } z = e^{k\pi i}= 1\textrm{)}\\ =36, & k \textrm{ odd} & \textrm{(so } z = e^{k\pi i}= -1\textrm{)} \end{cases}$$
Thus the magnitude of your original expression is minimized (with value $0$) at $z=1$, and is maximized (with value $6$) at $z=-1$.
On
My answer is very simple.Look:
$$|z^2-3z+2|=|(z-\frac{3}{2})^2-\frac{1}{4}|$$
We can write $$-1≤z≤1$$ $$-\frac{5}{2}≤z-\frac{3}{2}≤-\frac{1} {2}$$
Finally,
$$\frac{25}{4}≥(z-\frac{3}{2})^2≥\frac{1}{4}$$
$$6≥(z-\frac{3}{2})^2-\frac{1}{4}≥0$$
$$6≥|(z-\frac{3}{2})^2-\frac{1}{4}|≥0$$
$$f(z)_{max}=6, |z|≤1$$
Done.
The maximum of $z^2-3z+2$ is not defined because $z^2-3z+2\in\mathbb C$.
By the way, $$|z^2-3z+2|\leq|z|^2+3|z|+2\leq6.$$ The equality occurs for $z=-1$.