Given $|z^4 -8 + 3|, |z|=2 $
I need to find the positive upper and lower bounds of the above expression.
My attempt:
By triangle inequality, we have
$||z^4|+|-8z|+3|\geq|z^{4} -8z + 3| \leq ||z^4 -8z| +3|???||z^4|+|3-8z|| \geq ||z^4|-|8z|+3|\geq||z^4+3|-|8z|| $
(I used ??? since I don't know which is larger than which, and I believe i got all the combinations possible)
At the left most, I believe that will give me the upper bound so $|2^4|+|-8(2)|+3| = 35$
At the right most, $||2^4 +3| -|8(2)| = 3|$
BUT, and here is my question, given that $z = r*e^{i\theta}, |e^{i\theta}| = 1$
assuming my triangle inequality is true, we know $|8z| = |8r*e^{i\theta}|=|8r|$, however $|z^4+3|$ is not necessarily equal to $|r^4 +3|$, I wonder if $|z^4+3| = |8z|$ such that $||z^4+3|-|8z||= 0$ ?
Alternatively if $|z^4 - 8z| = -3$ is possible??
$\endgroup$ 122 Answers
$\begingroup$Given two complex numbers $a$ and $b$, then the triangle inequality is $$||a|-|b|| \le |a -b| \le |a|+|b|$$
Hence, if we want simple bounds, assuming $|z|=2$, then we can do the following, $$13= |16-3| = ||-8z|-|3|| \le |-8z+3| \le |-8z|+|3| = 16+3 = 19$$ Hence, $|-8z+3|$ is bounded between $13$ and $19$ (notice that it could be $16$). Thus, we have $$0\le |16-|-8z+3||=||z^4|-|-8z+3||\le |z^4-8z+3| \le |z^4|+|8z|+3 = 35$$ So, you have lower bound $0$ and upper bound $35$.
If you want tighter bounds, you could solve the absolute value explicitly. That is, let $z=x+yi$, then assuming $|z|=2$ we have that $y^2 = 4 - x^2$, so we perform substitution, $$|(x+yi)^4-8(x+yi)+3| $$ $$= |(x^4-6x^2y^2+y^4-8x+3)+(4x^3y-4xy^3-8y)i|$$ $$=\sqrt{(x^4-6x^2y^2+y^4-8x+3)^2+(4x^3y-4xy^3-8y)^2}=\sqrt{48x^4-256x^3-192x^2+720x+617}$$ we can explicitly take the derivative of the polynomial ($192x^3-768x^2-384x+720$) in the square root and find the absolute maximum and absolute minimum, where $x$ is between $-2$ and $2$. Doing so we find the absolute minimum is $1.79964$ and is found with $x=-1.0785$ ($z=-1.0785+1.68429i$), likewise the absolute maximum is $35$ and is found with $x=-2$ ($z=-2+0i$). So we have exact bounds, $$1.79964 \le |z^4-8z+3| \le 35$$ moreover equality is achieved at specific $z=-1.0785+1.68429i$ and $z=-2$.
$\endgroup$ 3 $\begingroup$A point on the circle is characterized by $z=2e^{i\theta}$.
Let $g(z)=z^4 - 8z+3.$
We seek the minimum and maximum of
$\begin{aligned} |g(\theta)|^2 &= g(2e^{i\theta})\overline{ g(2e^{i\theta})}=g(2e^{i\theta})g(2e^{-i\theta})\\&=(16e^{4i\theta}-16e^{i\theta}+3)(16e^{-4i\theta}-16e^{-i\theta}+3)\end{aligned}$
Differentiatate with respect to $\theta$ to find extrema $\cdots.$
$\endgroup$
