Question

The equation ${{z}^{3}}+iz-1=0$ has
(a) 3 real roots
(b) One real root
(c) No real roots
(d) No real/complex roots

Answer 1

Hint: For the equation consider that x is a real root. So put x in place of x. Separate them in the form of a complex number $\left( a+ib \right)$. Then equate the real and imaginary part to zero. Find the real number which satisfies both the equations.

Complete step-by-step answer:
We have been given an equation, for which we need to find the roots.
Given to us the expression,
${{z}^{3}}+iz-1=0$
Now let us suppose that x be the real root. Hence we can put x in the place of z in the above equation. Thus it becomes,
${{x}^{3}}-ix-1=0-(1)$
We know that a complex number is of the form $a+ib$, where a is a real number and b is imaginary number. Thus we can make (1) in the form of $a+ib$.
$\left( {{x}^{3}}-1 \right)-ix=0$
Now, $\left( {{x}^{3}}-1 \right)$ is the real part and (-x) is the imaginary part. For better understanding, we can write it as,
$\left( {{x}^{3}}-1 \right)-ix=0+i0$
Thus equate the real parts and imaginary part to zero.
i.e. ${{x}^{3}}-1=0$ and x = 0
From this we can say that, $x=\sqrt[3]{1}=1$ and x = 0. Hence, there is no real number that satisfies these two equations. Hence our assumption is wrong. Thus we can say that, ${{z}^{3}}+iz-1=0$ has no real roots.
Thus there are no real roots.
$\therefore$ Option (c) is the correct answer.

Note: We got the equation as ${{x}^{3}}-1=0$ and x = 0. There is no real number that can satisfy both these equations. When we solved it we got x = 0 and x = 1. But we didn’t get any real numbers that satisfy them.