by repeated applications of the triangle inequality
$\left|z^3-2z^2+6z+2\right| \leq |z^3| + |2z^2|+|6z|+|2| = |z|^3 + 2 |z|^2 + 6|z| + 2$
and if $|z|< 2$ then
$\left|z^3-2z^2+6z+2\right| < 2^3 + 2\cdot 2^2 + 6\cdot 2 + 2 = 30$
I am trying to figure out the following problem:
Find an upper bound on |z^{3}-2z^{2}+6z+2| defined by |z|<2.
I've done the following:
|z^{3}-2z^{2}+6z+2|
= |(x+iy)^{3}-2(x+iy)^{2}+6(x+iy)+2|
= |x^{3}+3ix^{2}y-3xy^{2}-iy^{3}-2(x^{2}+2ixy-y^{2})+6x+6iy+2|
= |x^{3}+3ix^{2}y-3xy^{2}-iy^{3}-2x^{2}-4ixy+2y^{2}+6x+6iy+2|
= |x^{3}-3xy^{2}-2x^{2}+2y^{2}+6x+2+(3x^{2}y-y^{3}-4xy+6y)i|
= √(x^{3}-3xy^{2}-2x^{2}+2y^{2}+6x+2)^{2}+ (3x^{2}y-y^{3}-4xy+6y)^{2 }I don't think I'm going this the right way. Any ideas? Should I not change to x+iy form?
by repeated applications of the triangle inequality
$\left|z^3-2z^2+6z+2\right| \leq |z^3| + |2z^2|+|6z|+|2| = |z|^3 + 2 |z|^2 + 6|z| + 2$
and if $|z|< 2$ then
$\left|z^3-2z^2+6z+2\right| < 2^3 + 2\cdot 2^2 + 6\cdot 2 + 2 = 30$