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 $2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left|  2z^2 \right|)= 2^{\frac {1}{2}}(\frac {3}{4} + \frac {9}{2})  = \frac {21}{2*2^{\frac {1}{2}}} \leq 9.5$

Credit: @MartinBladt is right that something is wrong with the question.
Let $z = \frac{\sqrt{5}}{2}+i$
$$|z|=\sqrt{\frac54+1}=\frac32$$
$$|\Im(2+z^c+-4z^2)|=|-\Im (z)-4(2) \Re(z)\Im(z)|=1+8\frac{\sqrt{5}}{2}=1+4\sqrt{5}>9.5$$
Edit:
The following statement is not valid: 
$$\left| z^{c} - 4z^2 \right|= 2^{\frac {1}{2}}\left| \frac {z^{c}}{2} - 2z^2 \right|  $$
In particular, we can let $z=1$, we can see that the left hand side is rational but the right hand side is irrational.
We do have the following equation: 
$$\left| z^{c} - 4z^2 \right|= 2\left| \frac {z^{c}}{2} - 2z^2 \right|  $$
