Question

If ${z_1}$, ${z_2}$ and ${z_3}$ are complex numbers such that $|{z_1}| = |{z_2}| = |{z_3}| = \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}}} \right| = 1$ then $|{z_1} + {z_2} + {z_3}|$equal
A. equals 1
B. less than 1
C. greater than 1
D. equals 3

Answer 1

Here The question is related to the complex number. In this question they have some conditions and these conditions are not general conditions. By using these conditions we have to find the value of the $|{z_1} + {z_2} + {z_3}|$. So we use the modulus identities i.e., $z\overline z = |z{|^2}$, $\overline {{z_1}} + \overline {{z_2}} = \overline {{z_1}{z_2}}$ and $|\overline {{z_1}} | = |{z_1}|$, we are obtaining the value of $|{z_1} + {z_2} + {z_3}|$.

Complete step by step answer:
The $z$ will represent the complex number. It is represented as $z = x + iy$. Here in this question some conditions are mentioned.. The given conditions are
$|{z_1}| = |{z_2}| = |{z_3}| = 1$ and $\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}}} \right| = 1$
Using these conditions we have to find the value of $|{z_1} + {z_2} + {z_3}|$. Now we consider,
$|{z_1}| = |{z_2}| = |{z_3}| = 1$
On squaring each and every term which is present in the above inequality.
$\Rightarrow |{z_1}{|^2} = |{z_2}{|^2} = |{z_3}{|^2} = {1^2}$

The number 1 square is 1. So we have
$\Rightarrow |{z_1}{|^2} = |{z_2}{|^2} = |{z_3}{|^2} = 1$
By the modulus identity of the complex number $z\overline z = |z{|^2}$, the inequality is written as
$\Rightarrow {z_1}\overline {{z_1}} = {z_2}\overline {{z_2}} = {z_3}\overline {{z_3}} = 1$
All the terms are equal to 1, so we equate for each term and it is written as
$\Rightarrow {z_1}\overline {{z_1}} = 1\,\,$, ${z_2}\overline {{z_2}} = 1$ and ${z_3}\overline {{z_3}} = 1$
Take ${z_1}$, ${z_2}$ and ${z_3}$ to the RHS we have
$\Rightarrow \overline {{z_1}} = \dfrac{1}{{{z_1}}}\,\,$, $\overline {{z_2}} = \dfrac{1}{{{z_2}}}$ and $\overline {{z_3}} = \dfrac{1}{{{z_3}}}$--------- (1)

Now we consider the another condition which is given in the question
$\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}}} \right| = 1$
By considering the equation the above inequality is written as
$\Rightarrow \left| {\overline {{z_1}} + \overline {{z_2}} + \overline {{z_3}} } \right| = 1$
As we know that the modulus identity $\overline {{z_1}} + \overline {{z_2}} = \overline {{z_1}{z_2}}$, using this the above inequality is written as
$\Rightarrow \left| {\overline {{z_1} + {z_2} + {z_3}} } \right| = 1$
Since $|\overline {{z_1}} | = |{z_1}|$, we have
$\therefore |{z_1} + {z_2} + {z_3}| = 1$
Therefore the value of $|{z_1} + {z_2} + {z_3}|$ is 1.

Hence option A is correct.

Note: While solving the problems related to the modulus, the most important thing we have to know about the modulus identities. Because these identities will make the problem much easier. Some modulus identities are:
1. $|{z_1}\;{z_2}\left| {{\text{ }} = {\text{ }}} \right|{z_1}\left| {\text{ }} \right|{z_2}|$
2. $\left| {\dfrac{{{z_{1\;}}}}{{{z_2}}}} \right|{\text{ }} = \dfrac{{\left| {{z_{1\;}}} \right|}}{{\left| {{z_2}} \right|}}{\text{ }}$
3. $\overline {{z_1}{z_2}} = \overline {{z_1}} \overline {{z_2}} \;$
4. ${z_1} \pm {z_2} = \overline {{z_1}} \pm \overline {{z_2}}$
5. $\overline {\left( {\dfrac{{{z_1}}}{{{z_2}}}} \right)} = \dfrac{{\overline {{z_1}} }}{{\overline {{z_2}} \;\;}}$, where ${z_2}\; \ne {\text{ }}0$
These identities are important and we have used some identities wherever it is necessary.