Q:=> \texttt{Prove that for all z with |z|=1, the relation given is true}}
\sqrt{2} \leq |1-z|+|1+z^{2}| \leq 4
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1 Answers
21think geometrically
proving the RHS
write the given expression as |z-1| + |1+ z2| ≤ |z-1| + |1| + |z2| (equality can hold)
|z-1| + |1| + |z2| = 2 + |z-1|≤ 4 because |z-1|≤2 (say graphically by finding the max distance between z and (1,0) where z lies in a circle of unit radious or simply use modulus inequality)
so we get given expression less than equal to 4
