<s>[INST] Triangle inequality in C Show that $\left|Im(2+ z^{c} -4z^2) \right| \leq 9.5$  When $ \left| z \right| \leq \frac {3}{2}$
$z^c$= compliment of z
$\left|Im(2+ z^{c} -4z^2) \right| \leq \left| 2+z^{c} - 4z^2 \right|$
I have tried to split it up directly i have tried to force complete the square i always get a weird value or a number bigger than 9.5.
EDIT: Why cant i write 
$\left| Im(2+ z^{c} -4z^2) \right| = \left| Im(z^{c} -4z^2)\right| \leq \left| z^{c} - 4z^2 \right|= 2^{\frac {1}{2}}\left| \frac {z^{c}}{2} - 2z^2 \right|  \leq 2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left|  2z^2 \right|)$
and $2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left|  2z^2 \right|)= 2^{\frac {1}{2}}(\frac {3}{4} + \frac {9}{2})  = \frac {21}{2*2^{\frac {1}{2}}} \leq 9.5$
 [/INST] You're trying to prove the unprovable. Let $z=re^{it}$, such that 
$$Im(2+ z^{c} -4z^2) =Im(z^c)-4Im(z^2)=-r\sin (t)-4r^2sin(2t).$$
For $r=3/2$, this expression is $\approx-10.076$ at $t\approx 0.814$.
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