Question

If the fourth roots of unity are \[{z_1},{z_2},{z_3},{z_4}\]
then ${z_1}^2 + {z_2}^2 + {z_3}^2 + {z_4}^2$ is equal to
A.$1$
B.$0$
C.$i$
D.None of these

Answer 1

Hint :- Make use of the concept of fourth roots of unity and solve this
Fourth roots of Unity
Properties of Four Fourth Roots of Unity
a. Sum of all the four fourth roots of unity is zero.
b. The real fourth roots of unity are additive
    Inverse of each other.
c. Both the complex / imaginary Fourth roots of
    unity are conjugate for each other
d. Product of all the Fourth roots of unity is –


Complete step by step by solution
Let $x$ be the four fourth roots of $1$, if then we can write
$x = 4\sqrt 1 $
We should write it
$x = {(1)^{\dfrac{1}{4}}}$
$ \Rightarrow {x^4} = 1$
$ \Rightarrow {x^4} - {1^4} = 0$
\[ \Rightarrow {({x^2})^2} - {({1^2})^2} = 0\]
\[[{a^2} - {b^2} = (a + b)(a - b)]\]
Therefore,
\[ \Rightarrow ({x^2} - 1)({x^2} + 1) = 0\]
Either,
\[({x^2} - 1) = 0 or ({x^2} + 1) = 0\]
\[{x^2} = 1 or {x^2} = - 1\]
\[x = \pm \sqrt 1 or x = \pm \sqrt { - 1} \]
\[x = \pm 1 or x = \pm i\]
Now, the Four fourth roots are unity is $[1, - 1,i, - i]$
Now we complete the answer
Step by step
(Image)
\[{z_1},{z_2},{z_3},{z_4}\] are roots of
${x^4} - 1 = 0$
\[\therefore {z_1} + {z_2} + {z_3} + {z_4} = 0\]
\[{z_1}{z_2} + {z_2}{z_3} + {z_3}{z_4} + {z_4}{z_1} + {z_1}{z_3} + {z_2}{z_4} = 0\]
\[\therefore {({z_1} + {z_2} + {z_3} + {z_4})^2} = \sum\limits_{}^{} {{z_1}^2} \]
$\sum\limits_{i = 1}^4 {} \sum\limits_{i = 1}^4 {} {z_1}{z_i}$
\[0 = {\sum {{z_1}} ^2} = 0\]
\[\therefore {\sum {{z_1}} ^2} = 0\]

So B is the Answer
B=$0$

Note–Complex numbers are the numbers which are expressed in the form of $a + ib$, where $i$ is an imaginary number called iota and has the value of $\sqrt { - 1} $.
Therefore, the combination of both real and imaginary numbers is a complex number.