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

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    Let $z=r(\cos t+i\sin t)$ Im$(2+z^c-4z^2)=-r\sin t-4r^2\sin2t$ – 2017-01-20
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    It very clever but trying to say something about the length of sin t is difficult i know its at most 1 but that gives 10.5 $\leq$ 9.5 maybes its 2 - 1.5 - 9 = 8.5? – 2017-01-20

2 Answers 2

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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:

$$\left| z^{c} - 4z^2 \right|= 2\left| \frac {z^{c}}{2} - 2z^2 \right| $$

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    Why doesnt my edit work im not breaking any rules of algebra normaly $( (ab)^2 +b^2) ^{\frac {1}{2}} = b ((a)^2 +1^2) ^{\frac {1}{2}}$ – 2017-01-20
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    pointed out a mistake in one of the equation in the edit. – 2017-01-20
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    Correcting the mistake from $\sqrt{2}$ to $2$ does give you an upperbound of $10.5$. – 2017-01-20
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    Hmm question must be wrong on explanation thanks for all the work still not sure why i cant factor out a 2. but ok ^^ – 2017-01-20
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    For a complex number, if we multiply it by $2$, the length would be multiplied by $2$ rather than $\sqrt{2}$. – 2017-01-20
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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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    where did the +2 go? – 2017-01-20
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    $2$ is a real number, $\Im(2)=0$. – 2017-01-20
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    then we should use that Im|z| $\leq |z|$ to keep the +2 – 2017-01-20
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    I come out with an explicit simple example in case there is any computational mistake. Do you agree that the upper bound is not valid? – 2017-01-20