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Question: If log<sub>1/2</sub>\(\frac{|z|^{2} + 2|z| + 4}{2|z|^{2} + 1}\)\< 0 then the region...

If log1/2∣z∣2+2∣z∣+42∣z∣2+1\frac{|z|^{2} + 2|z| + 4}{2|z|^{2} + 1}< 0 then the region

A

1 < | z | < 3

B

| z | < 2

C

| z | < 3

D

| z | > 1

Answer

| z | < 3

Explanation

Solution

Sol. log⁡1/2∣z∣2+2∣z∣+42∣z∣2+1<0\log_{1/2}\frac{|z|^{2} + 2|z| + 4}{2|z|^{2} + 1} < 0

∣z∣2+2∣z∣+42∣z∣2+1>(12)0\frac{|z|^{2} + 2|z| + 4}{2|z|^{2} + 1} > \left( \frac{1}{2} \right)^{0}

∣z∣2−2∣z∣−3|z|^{2} - 2|z| - 3 < 0

(| z | – 3) (| z | + 1) < 0

| z | < 3