Question

Find the number of real or purely imaginary solution of the equation ${z^3} + iz - 1 = 0$ is:
A.Zero
B.One
C.two
D.three

Answer 1

Hint : A complex number is the number generally expressed in the form of $a + ib$ .
In this question we are given with a complex equation so we will first substitute the value of the complex number and then will solve the equation and by substituting the value will separate the imaginary part and the real part in the equation and by substituting the value of real and imaginary numbers will find the number of solutions.

Complete step-by-step answer :
Given the equation whose solution is to be found is ${z^3} + iz - 1 = 0$ , here $z$ is the complex number
Let us assume that the complex number $z = x + iy$ where $x,y \in R$ , since a complex number is expressed in the form of $a + ib$
Now we substitute $z = x + iy$ in the given equation, we get
${\left( {x + iy} \right)^3} + i\left( {x + iy} \right) - 1 = 0$
By solving this equation we get

{x^3} + 3\left( {{x^2}} \right)\left( {iy} \right) + 3x{\left( {iy} \right)^2} + {\left( {iy} \right)^3} + ix + {i^2}y - 1 = 0 \\\ {x^3} + 3\left( {{x^2}} \right)\left( {iy} \right) - 3x{y^2} - i{y^3} + ix - y - 1 = 0\left\\{ {\because {i^2} = - 1} \right\\} \\\ \left( {{x^3} - 3x{y^2} - y - 1} \right) + i\left( {3{x^2}y - {y^3} + x} \right) = 0 \;

Hence we get real part of the equation as $\left( {{x^3} - 3x{y^2} - y - 1} \right)$ and imaginary part as $\left( {3{x^2}y - {y^3} + x} \right)$
Now we can write $0 + i0 = 0$ , hence the equation can be written as $\left( {{x^3} - 3x{y^2} - y - 1} \right) + i\left( {3{x^2}y - {y^3} + x} \right) = 0 + i0$
Equate real part and imaginary part in the equation, we get

$$\left( {{x^3} - 3x{y^2} - y - 1} \right) = 0 - - (i) \\\ \left( {3{x^2}y - {y^3} + x} \right) = 0 - - (ii) \;$$

Now let’s find the solution of both the equation by substituting $x = 0$ in both the equation, we get

$$\left( {{x^3} - 3x{y^2} - y - 1} \right) = 0 \\\ \left( {{{\left( 0 \right)}^3} - 3\left( 0 \right){y^2} - y - 1} \right) = 0 \\\ \Rightarrow - y - 1 = 0 \\\ \& \\\ \left( {3{{\left( 0 \right)}^2}y - {y^3} + 0} \right) = 0 \\\ \Rightarrow - {y^3} = 0 \;$$

Now since we get y = 0\left\\{ {y \in R} \right\\} , we can say a solution is not possible.
Now the solution, when $y = 0$ in both the equation, we get

$$\left( {{x^3} - 3x{{\left( 0 \right)}^2} - \left( 0 \right) - 1} \right) = 0 \\\ \Rightarrow {x^3} - 1 = 0 \\\ \& \\\ \left( {3{x^2}\left( 0 \right) - {{\left( 0 \right)}^3} + x} \right) = 0 \\\ \Rightarrow x = 0 \;$$

Now since we get x = 0\left\\{ {x \in R} \right\\} , we can say a solution is not possible.
Therefore, zero solution is possible for the equation.
Option A is correct.
So, the correct answer is “Option A”.

Note : To find the solution of a complex equation the real part of the equation can only be equated to the real part of the solution and the imaginary part is equated with the imaginary part of the solution. In a complex number $a + ib$ , $a$ and $b$ is the real number, $ib$ is the imaginary part and $i$ is the imaginary unit.