The value of $\dfrac{{{i}^{592}}+{{i}^{590}}+{{i}^{588}}+{{i}^{586}}+{{i}^{584}}}{{{i}^{582}}+{{i}^{580}}+{{i}^{578}}+{{i}^{576}}+{{i}^{574}}}-1=$
a). -1
b). -2
c). -3
d). -4

Hint: We are going to use the value of power of iota (i), $ i = i, i^2 = -1, i^3 = i^2(i) = -i, i^4 = i^2(i^2) = 1 $. So, divide the power of iota (i) given in the question by 4 then different remainders have different values. Then simplify the expression by substituting these “i” values in the given expression.

Complete step-by-step solution -
We know that from the complex number that:
$\begin{align}
  & i=i \\
 & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
 & {{i}^{4}}=1 \\
\end{align}$
Now, we can convert these different powers of i into the general expression of the power in iota (i).
$\begin{align}
  & {{i}^{4n}}=1 \\
 & {{i}^{4n+1}}=i \\
 & {{i}^{4n+2}}=-1 \\
 & {{i}^{4n+3}}=-i \\
 & \\
\end{align}$
In the first expression of the above iota (i) equations, the power of iota (i.e. $ i^{4n} $) is divisible by 4. In the second expression, on dividing the power of iota (i.e. $ i^{4n+1}$) by 4 we get 1. In the third expression, the power of iota (i.e. $i^{4n+2}$) when divided by 4 gives 2 and in the last expression, the power of iota (i.e. $i^{4n+3}$) when divided by 4 gives 3.
Now, we can use these different powers of iota (i) in the above expression and then simplify.
$\dfrac{{{i}^{592}}+{{i}^{590}}+{{i}^{588}}+{{i}^{586}}+{{i}^{584}}}{{{i}^{582}}+{{i}^{580}}+{{i}^{578}}+{{i}^{576}}+{{i}^{574}}}-1$
We are going to divide the power of iota (i) by 4.
$\dfrac{{{i}^{4}}+{{i}^{2}}+{{i}^{4}}+{{i}^{2}}+{{i}^{4}}}{{{i}^{2}}+{{i}^{4}}+{{i}^{2}}+{{i}^{4}}+{{i}^{2}}}-1$
Substituting the values of $ i^4 = 1 $ and $ i^2 = -1 $ in the above expression we get,
$\begin{align}
  & \dfrac{1-1+1-1+1}{-1+1-1+1-1}-1 \\
 & \Rightarrow \dfrac{1}{-1}-1 \\
 & \Rightarrow -1-1 \\
 & \Rightarrow -2 \\
\end{align}$
Hence, the simplification of the expression gives the value -2.
Hence, the correct option is (b).

Note: While dividing the power of iota (i) by 4, you can use the divisibility test of 4. Instead of dividing the huge powers of iota (i), you can just divide the last two numbers of the power and can see when dividing the last two numbers what remainder we are getting. This gives the same result as the one by dividing the whole number by 4.