Question

If \[z_1\] and \[z_2\] are the complex roots of the equation \[{\left( {x - 3} \right)^3} + 1 = 0\] then the value of \[z_1 + z_2\] is
A. 6
B. 3
C. 5
D. 7

Answer 1

Hint: Here first we will find the roots of the given equation and then select the complex roots from the 3 roots obtained and then add them to get the desired value.

Complete step-by-step answer:
The given equation is:
\[{\left( {x - 3} \right)^3} + 1 = 0\] and \[z_1\] and \[z_2\] are the complex roots of this equation.
Solving the given equation we get:-
\[
  {\left( {x - 3} \right)^3} + 1 = 0 \\
  {\left( {x - 3} \right)^3} = - 1 \\
  \dfrac{{{{\left( {x - 3} \right)}^3}}}{{ - 1}} = 1 \\
 \]
Since we know that:
\[{\left( { - 1} \right)^3} = - 1\]
Therefore the above equation can be written as:
\[{\left( {\dfrac{{x - 3}}{{ - 1}}} \right)^3} = 1\]
Now taking cube root both the sides we get:
\[\dfrac{{x - 3}}{{ - 1}} = \sqrt[3]{1}\]
Now as we know that cube root of unity is equal to:
\[\sqrt[3]{1} = 1,w,{w^2}\]
Therefore,
\[\dfrac{{x - 3}}{{ - 1}} = 1,\dfrac{{x - 3}}{{ - 1}} = w,\dfrac{{x - 3}}{{ - 1}} = {w^2}\]
Now solving the above equations for x we get:
\[
  x - 3 = - 1 \\
  x = 3 - 1 \\
  x = 2; \\
  x - 3 = - w \\
  x = 3 - w; \\
  x - 3 = - {w^2} \\
  x = 3 - {w^2} \\
 \]
So we get three values of x:
\[x = 1,x = 3 - w,x = 3 - {w^2}\]
Now since we know that \[x = 1\] is a real root and \[x = 3 - w,x = 3 - {w^2}\] are the complex roots therefore,
\[
  z_1 = 3 - w \\
  z_2 = 3 - {w^2} \\
 \]
Now adding both the complex roots we get:
\[
  z_1 + z_2 = 3 - w + 3 - {w^2} \\
  z_1 + z_2 = 6 - w - {w^2} \\
 \]
Using the following formula:
\[
  1 + w + {w^2} = 0 \\
   - w - {w^2} = 1 \\
 \]
We get:
\[
  z_1 + z_2 = 6 + 1 \\
  z_1 + z_2 = 7 \\
 \]
Hence the desired value is 7 and therefore option D is correct.

Note: Make a note that the cube roots of unity are \[1,w,{w^2}\]. Also, \[1 + w + {w^2} = 0\].