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

Hint:
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 \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].