Question

If $\left| {{z}_{1}} \right|=1,\left| {{z}_{2}} \right|=2,\left| {{z}_{3}} \right|=3$ and $\left| 9{{z}_{1}}{{z}_{2}}+4{{z}_{1}}{{z}_{3}}+{{z}_{2}}{{z}_{3}} \right|=12$ then the value of $\left| {{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right|$ is equal to ?
a)2
b)3
c)4
d)6

Answer 1

Hint: We will use two formula to solve this question $z\overline{z}={{\left| z \right|}^{2}}$ also, $\overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}}=\overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}$. First we will solve the given relation $\left| 9{{z}_{1}}{{z}_{2}}+4{{z}_{1}}{{z}_{3}}+{{z}_{2}}{{z}_{3}} \right|=12$, to get the value of $\left| \overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}} \right|$ and then from this we will find the value of $\left| {{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right|$, using the second relation.

Complete step-by-step answer:

It is given in the question that $\left| {{z}_{1}} \right|=1,\left| {{z}_{2}} \right|=2,\left| {{z}_{3}} \right|=3$ also a relation is given - $\left| 9{{z}_{1}}{{z}_{2}}+4{{z}_{1}}{{z}_{3}}+{{z}_{2}}{{z}_{3}} \right|=12$ and then we have to find the value of $\left| \overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}} \right|$. Now, we know that for any complex number z, $z\overline{z}={{\left| z \right|}^{2}}$ thus, from the formula we get, ${{z}_{1}}\overline{{{z}_{1}}}={{\left| 1 \right|}^{2}}=1$ and ${{z}_{2}}\overline{{{z}_{2}}}={{\left| 2 \right|}^{2}}=4$ and ${{z}_{3}}\overline{{{z}_{3}}}={{\left| 3 \right|}^{2}}=9$. Now, we will replace the 9 as ${{z}_{3}}\overline{{{z}_{3}}}$ , 4 as ${{z}_{2}}\overline{{{z}_{2}}}$ and 1 as \[{{z}_{1}}\overline{{{z}_{1}}}\] in the given relation $\left| 9{{z}_{1}}{{z}_{2}}+4{{z}_{1}}{{z}_{3}}+{{z}_{2}}{{z}_{3}} \right|=12$,
We get - $\left| {{z}_{3}}\overline{{{z}_{3}}}{{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{3}}{{z}_{2}}\overline{{{z}_{2}}}+{{z}_{2}}{{z}_{3}}{{z}_{1}}\overline{{{z}_{1}}} \right|=12$ taking ${{z}_{1}}\centerdot {{z}_{2}}\centerdot {{z}_{3}}$ common in LHS we get - $\left| {{z}_{1}}\centerdot {{z}_{2}}\centerdot {{z}_{3}} \right|\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right|=12$, Solving further, we get $\left| {{z}_{1}} \right|\centerdot \left| {{z}_{2}} \right|\centerdot \left| {{z}_{3}} \right|\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right|=12$ we have ${{z}_{1}}=1,{{z}_{2}}=2$ and ${{z}_{3}}=3$, on putting these values in the above equation we get $1\cdot 2\cdot 3\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right|=12$, that is, $6\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right|=12$ or $\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right|=\dfrac{12}{6}=2$.
Now, we know that $\overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}}=\overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}$, therefore we can say that \[\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}} \right|=\left| \overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}} \right|=2\] or we get the value of expression $\left| \overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}} \right|=2$ thus option a) is the correct answer.

Note: Usually student get stuck in the last step because most of the student don’t know that the value $\overline{{{z}_{1}}}+\overline{{{z}_{2}}}+\overline{{{z}_{3}}}=\overline{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}$ holds true. And they get stuck just before a few steps to finish their answer. Thus it is recommended to learn all the properties of complex numbers to solve this type of problem easily and completely.