Question

If z lies on the curve |z| = 1 such that a $\le$ |z+1| + |1 + $z^2$ - z| $\le$ b then (a,b) can be

• A. (1 ,3)
• B.
• C.
• D.

Answer 1

The correct option is C

Let x = |z+1|

x = |z+1| $\le |z|+1=2\Rightarrow x\in [0,2]$

$x^2=|z+1{|}^2=(z+1)(\overline{z}+1)$

= 2 + z + $\overline{z}$ $\Rightarrow z+\overline{z}=x^2-2$

now $|1+z^2-z|=|z\overline{z}+z^2-z|=|\overline{z}+z-1|$

= $|x^2-2-1|=|x^2-3|=3-x^2$

f(x) = |z+1| + $|1+z^2-z|$ = $-x^2+x+3$

$f_{max}\left(x\right)=\frac{4.(-1).3-1}{4(-1)}=\frac{13}{4}atx=\frac{1}{2}$

$f_{min}\left(x\right)=f\left(2\right)=1$

Checking for extreme values gives required range of f(x) as $[1,\frac{13}{4}]$.