Question

If $|{z_1}| = |{z_2}| = |{z_3}| = ...... = |{z_n}| = 1$ , then prove that $|{z_1} + {z_2} + {z_3} + ..... + {z_n}| = \left| {\dfrac{1}{{{z_1}}}} \right. + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}} + ..... + \left. {\dfrac{1}{{{z_n}}}} \right|$.

Answer 1

Hint:For solving this particular problem we first take the left hand side of the expression then try to convert it in the form of the expression present in the right side of the equation. We can convert the left side by multiplying both numerator and denominator by the conjugate of the corresponding complex numerator, then use ${\left| {{z_1}} \right|^2} = {z_1}\overline {{z_1}} $ , for simplifying the expression use the given expression that is $|{z_1}| = |{z_2}| = |{z_3}| = ...... = |{z_n}| = 1$ . After evaluating and substituting we get our required solution.

Formula Used:
For solving this particular solution ,we used the following relationship,
${\left| {{z_1}} \right|^2} = {z_1}\overline {{z_1}} $ , here \[\overline {{z_1}} \] means the conjugate of ${z_1}$.

Complete step by step answer:
We have given that,$|{z_1}| = |{z_2}| = |{z_3}| = ...... = |{z_n}| = 1$
To prove: $|{z_1} + {z_2} + {z_3} + ..... + {z_n}| = \left| {\dfrac{1}{{{z_1}}}} \right. + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}} + ..... + \left. {\dfrac{1}{{{z_n}}}} \right|$
Proof :Taking L.H.S of the given expression that is ,
$|{z_1} + {z_2} + {z_3} + ..... + {z_n}|$
Multiplying both numerator and denominator by the conjugate of the corresponding complex numerator,
$\left| {\dfrac{{{z_1}\overline {{z_1}} }}{{\overline {{z_1}} }}} \right. + \dfrac{{{z_2}\overline {{z_2}} }}{{\overline {{z_2}} }} + \dfrac{{{z_3}\overline {{z_3}} }}{{\overline {{z_3}} }} + ..... + \left. {\dfrac{{{z_n}\overline {{z_n}} }}{{\overline {{z_n}} }}} \right|$
Now we know that ${\left| {{z_1}} \right|^2} = {z_1}\overline {{z_1}}$, multiplication of the complex number with the conjugate of the complex number we get magnitude which represents the distance of the complex number from the origin.
Therefore ,
$\left| {\dfrac{{{z_1}\overline {{z_1}} }}{{\overline {{z_1}} }}} \right. + \dfrac{{{z_2}\overline {{z_2}} }}{{\overline {{z_2}} }} + \dfrac{{{z_3}\overline {{z_3}} }}{{\overline {{z_3}} }} + ..... + \left. {\dfrac{{{z_n}\overline {{z_n}} }}{{\overline {{z_n}} }}} \right| = \left| {\dfrac{{{{\left| {{z_1}} \right|}^2}}}{{\overline {{z_1}} }}} \right. + \dfrac{{{{\left| {{z_2}} \right|}^2}}}{{\overline {{z_2}} }} + \dfrac{{{{\left| {{z_3}} \right|}^2}}}{{\overline {{z_3}} }} + ..... + \left. {\dfrac{{{{\left| {{z_n}} \right|}^2}}}{{\overline {{z_n}} }}} \right|.............(A)$
Now we know that,
$|{z_1}| = |{z_2}| = |{z_3}| = ...... = |{z_n}| = 1$ (given)
Now Squaring the given expression, we get the following ,
$|{z_1}{|^2} = |{z_2}{|^2} = |{z_3}{|^2} = ...... = |{z_n}{|^2} = {1^2}......(1)$
Now substitute the value of $(1)$ in $(A)$ ,
$\left| {\dfrac{{{{\left| {{z_1}} \right|}^2}}}{{\overline {{z_1}} }}} \right. + \dfrac{{{{\left| {{z_2}} \right|}^2}}}{{\overline {{z_2}} }} + \dfrac{{{{\left| {{z_3}} \right|}^2}}}{{\overline {{z_3}} }} + ..... + \left. {\dfrac{{{{\left| {{z_n}} \right|}^2}}}{{\overline {{z_n}} }}} \right| \\
\Rightarrow\left| {\dfrac{1}{{\overline {{z_1}} }}} \right. + \dfrac{1}{{\overline {{z_2}} }} + \dfrac{1}{{\overline {{z_3}} }} + ..... + \left. {\dfrac{1}{{\overline {{z_n}} }}} \right| \\
\Rightarrow \overline {\left| {\dfrac{1}{{{z_1}}}} \right. + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}} + ..... + \left. {\dfrac{1}{{{z_n}}}} \right|} \\
\therefore \left| {\dfrac{1}{{{z_1}}}} \right. + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}} + ..... + \left. {\dfrac{1}{{{z_n}}}} \right|$
Hence, Proved.

Note:As we know that $z = x + yi$ , which is the representation of the complex number. And $z = x - yi$ , is the conjugate of the complex number.Now multiplication of the complex number with the conjugate of the complex number we get magnitude which represents the distance of the complex number from the origin.