For the equation ${x^2} + {k^2} = \left( {2k + 2} \right)x,k \in R$, roots are complex, then
1. $k = - \dfrac{1}{2}$
2. $k > - \dfrac{1}{2}$
3. $k < - \dfrac{1}{2}$
4. $\left( { - \dfrac{1}{2}} \right) < k < 0$

Hint:In this question, we are given the equation ${x^2} + {k^2} = \left( {2k + 2} \right)x,k \in R$ and its roots are complex. The first step is to convert the equation into the general form of a quadratic equation. Now take the condition ${b^2} - 4ac < 0$ and calculate the value of $k$.

Formula Used:
The general form of the quadratic equation – The quadratic formula is a formula in elementary algebra that provides the solution to a quadratic equation. Other methods for solving a quadratic equation besides the quadratic formula include factoring, completing the square, graphing, and others.
$a{x^2} + bx + c = 0$
Quadratic formula –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step Solution:
Given that,
${x^2} + {k^2} = \left( {2k + 2} \right)x,k \in R$
${x^2} - \left( {2k + 2} \right)x + {k^2} = 0$
Compare the above equation with the general form of quadratic equation i.e., $a{x^2} + bx + c = 0$,
It implies that,
$a = 1,b = - \left( {2k + 2} \right),c = {k^2}$
Now, using the quadratic formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Also, given roots are complex
Therefore, ${b^2} - 4ac < 0$
${\left( { - \left( {2k + 2} \right)} \right)^2} - 4\left( 1 \right)\left( {{k^2}} \right) < 0$
$4{k^2} + 4 + 8k - 4{k^2} < 0$
$k < - \dfrac{1}{2}$


Hence, the correct option is 3.

Note: The key concept involved in solving this problem is good knowledge of complex quadratic equation roots. Students must know that imaginary numbers (and complex roots) appear in quadratic equations when the value under the radical portion of the quadratic formula is negative. When this happens, the equation in the set of real numbers has no roots (or zeros). The roots are complex numbers and will be referred to as "complex roots" (or "imaginary roots"). These complex roots will be expressed as $a + ib$.