Question

Let complex number $z$ be such that $\left| {z - \dfrac{6}{z}} \right| = 5$, then the maximum value of $\left| z \right|$ will be:
(a) 2
(b) 3
(c) 5
(d) 6

Answer 1

Here, we will use the concept of complex numbers. We will use the property of sum of complex numbers and factorisation of a quadratic polynomial by splitting the middle term to find the maximum value of $\left| z \right|$.

Formula Used: We will use the following formulas:

1. If a complex number $\left| z \right|$ is the sum of two complex numbers $\left| {{z_1}} \right|$ and $\left| {{z_2}} \right|$, then $\left| z \right| = \left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|$.
2. If $\left( {x - a} \right)\left( {x - b} \right) < 0$, where $a < b$, then $a < x < b$.

Complete step by step solution:
We will use the sum property of complex number to find the maximum value of $\left| z \right|$.
Substituting ${z_1} = z - \dfrac{6}{z}$ and ${z_2} = \dfrac{6}{z}$ in the formula $\left| z \right| = \left| {{z_1} + {z_2}} \right| \le \left| {{z_1}} \right| + \left| {{z_2}} \right|$, we get
$\Rightarrow \left| {z - \dfrac{6}{z} + \dfrac{6}{z}} \right| \le \left| {z - \dfrac{6}{z}} \right| + \left| {\dfrac{6}{z}} \right|$
Simplifying the expression, we get
$\Rightarrow \left| z \right| \le \left| {z - \dfrac{6}{z}} \right| + \dfrac{{\left| 6 \right|}}{{\left| z \right|}}$
Since 6 is a positive number, its absolute value is also 6.
Therefore, the equation becomes
$\Rightarrow \left| z \right| \le \left| {z - \dfrac{6}{z}} \right| + \dfrac{6}{{\left| z \right|}}$
Substituting $\left| {z - \dfrac{6}{z}} \right| = 5$ in the inequation, we get
$\Rightarrow \left| z \right| \le 5 + \dfrac{6}{{\left| z \right|}}$
Let $\left| z \right| = y$.
Rewriting the inequation, we get
$\Rightarrow y \le 5 + \dfrac{6}{y}$
Multiplying both sides of the inequation by $y$ using the distributive law of multiplication, we get
$\begin{array}{l} \Rightarrow y\left( y \right) \le \left( {5 + \dfrac{6}{y}} \right)\left( y \right)\\\ \Rightarrow {y^2} \le 5y + 6\end{array}$
Subtracting $5y$ and 6 from both sides of the inequation, we get
$\Rightarrow {y^2} - 5y - 6 \le 0$
This inequation is true for all values of $y$ that satisfy the quadratic polynomial ${y^2} - 5y - 6$.
We will factorise the quadratic polynomial by splitting the middle term.
Factorising the quadratic polynomial, we get
$\begin{array}{l} \Rightarrow {y^2} - 6y + y - 6 \le 0\\\ \Rightarrow y\left( {y - 6} \right) + 1\left( {y - 6} \right) \le 0\\\ \Rightarrow \left( {y - 6} \right)\left( {y + 1} \right) \le 0\end{array}$
We know that if $\left( {x - a} \right)\left( {x - b} \right) < 0$, where $a < b$, then $a < x < b$.
Therefore, since $\left( {y - 6} \right)\left( {y + 1} \right) \le 0$, we get
$\Rightarrow - 1 \le y \le 6$
Therefore, the maximum value of $y$ is 6.
Substituting $\left| z \right| = y$, we get the maximum value of $\left| z \right|$ as 6.

Thus, the correct option is option (d).

Note:
We multiplied both sides of the inequation $y \le 5 + \dfrac{6}{y}$ by $y$ without changing the sign of inequality. This is because since $\left| z \right|$ is always positive, $y$ is also positive.
We have used the distributive law of multiplication to multiply $\left( y \right)$ by $\left( {5 + \dfrac{6}{y}} \right)$. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$.