Question:
If $\left|z-\frac{3}{2}\right|=2$ , then the greatest value of $\left|z\right|$ is
If $\left|z-\frac{3}{2}\right|=2$ , then the greatest value of $\left|z\right|$ is
Updated On: May 18, 2024
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The Correct Option is C
Solution and Explanation
Given, $\left|z-\frac{3}{z}\right|=2$
$\because |z|=\left|\left(z-\frac{3}{z}\right)+\frac{3}{z}\right|$
$\Rightarrow |z| \leq\left|z-\frac{3}{z}\right|+\frac{3}{|z|}[\because|a+b| \leq|a|+|b|]$
$\Rightarrow |z| \leq 2+\frac{3}{|z|}$
$\Rightarrow |z|^{2} \leq 2|z|+3$
$\Rightarrow |z|^{2}-2|z|-3 \leq 0$
$\Rightarrow(|z|-3)(|z|+1) \leq 0$
$\Rightarrow -1 \leq|z| \leq 3$
Hence, the greatest value of $|z|$ is 3 .
$\because |z|=\left|\left(z-\frac{3}{z}\right)+\frac{3}{z}\right|$
$\Rightarrow |z| \leq\left|z-\frac{3}{z}\right|+\frac{3}{|z|}[\because|a+b| \leq|a|+|b|]$
$\Rightarrow |z| \leq 2+\frac{3}{|z|}$
$\Rightarrow |z|^{2} \leq 2|z|+3$
$\Rightarrow |z|^{2}-2|z|-3 \leq 0$
$\Rightarrow(|z|-3)(|z|+1) \leq 0$
$\Rightarrow -1 \leq|z| \leq 3$
Hence, the greatest value of $|z|$ is 3 .
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
