Let $i^2$ = $-1$, then $\left( {{i}^{10}}-\dfrac{1}{{{i}^{11}}} \right)+\left( {{i}^{11}}-\dfrac{1}{{{i}^{12}}} \right)+\left( {{i}^{12}}-\dfrac{1}{{{i}^{13}}} \right)+\left( {{i}^{13}}-\dfrac{1}{{{i}^{14}}} \right)+\left( {{i}^{14}}+\dfrac{1}{{{i}^{15}}} \right)$ is equal to
A. $-1$ + i
B. $-1$ - i
C. $1$ + i
D. –i
E. i

Hint: The numbers can be categorized into complex numbers and real numbers.
Complex numbers are imaginary numbers and are not used in usual calculations.

Complete step by step answer:
Complex numbers can be made by combination of both real numbers and imaginary numbers.
The addition of imaginary numbers can be imaginary or real and the same applies for other operations like subtraction, multiplication and division.
The basic unit of complex numbers is i. Letter “’i” represents the value as i = $\sqrt{-1}$ .
 When i is multiplied by itself, the value i2 is obtained.
${{i}^{2}}$ is derived as:
i × i = $\sqrt{-1}$ × $\sqrt{-1}$
${{i}^{2}}$ = ${{\left( \sqrt{-1} \right)}^{2}}$
${{i}^{2}}$ = $-1$
The expression whose value has to be determined is given as:
Expression = $\left( {{i}^{10}}-\dfrac{1}{{{i}^{11}}} \right)+\left( {{i}^{11}}-\dfrac{1}{{{i}^{12}}} \right)+\left( {{i}^{12}}-\dfrac{1}{{{i}^{13}}} \right)+\left( {{i}^{13}}-\dfrac{1}{{{i}^{14}}} \right)+\left( {{i}^{14}}+\dfrac{1}{{{i}^{15}}} \right)$
All imaginary numbers, i with even power will have the value $1$ as value of ${{i}^{2}}$ is equal to $-1$ and thereby each even power can be written as multiple of $2$ .
Expression =\[\begin{align}
  & \left( {{i}^{10}}-\dfrac{1}{{{i}^{11}}} \right)+\left( {{i}^{11}}-\dfrac{1}{{{i}^{12}}} \right)+\left( {{i}^{12}}-\dfrac{1}{{{i}^{13}}} \right)+\left( {{i}^{13}}-\dfrac{1}{{{i}^{14}}} \right)+\left( {{i}^{14}}+\dfrac{1}{{{i}^{15}}} \right) \\
 & =\left[ {{\left( {{\left( i \right)}^{2}} \right)}^{5}}-\dfrac{1}{{{i}^{10}}\times i} \right]+\left[ {{i}^{10}}\times i-\dfrac{1}{{{\left( {{(i)}^{2}} \right)}^{6}}} \right]+\left[ {{\left( {{\left( i \right)}^{2}} \right)}^{6}}-\dfrac{1}{{{i}^{12}}\times i} \right]+\left[ {{i}^{12}}\times i-\dfrac{1}{{{\left( {{(i)}^{2}} \right)}^{7}}} \right]+\left[ {{\left( {{\left( i \right)}^{2}} \right)}^{7}}+\dfrac{1}{{{i}^{14}}\times i} \right] \\
 & =\left[ {{\left( -1 \right)}^{5}}-\dfrac{1}{{{(-1)}^{5}}\times i} \right]+\left[ {{(-1)}^{5}}\times i-\dfrac{1}{{{(-1)}^{6}}} \right]+\left[ {{(-1)}^{6}}-\dfrac{1}{{{(-1)}^{6}}\times i} \right]+\left[ {{(-1)}^{6}}\times i-\dfrac{1}{{{\left( -1 \right)}^{7}}} \right]+\left[ {{\left( -1 \right)}^{7}}+\dfrac{1}{{{(-1)}^{7}}\times i} \right] \\
 & =\left( -1+\dfrac{1}{i} \right)+(-i-1)+\left( 1-\dfrac{1}{i} \right)+(i+1)+\left( -1-\dfrac{1}{i} \right) \\
 & =-1-\dfrac{1}{i} \\
\end{align}\]
As ${{i}^{2}}$ = \[-1\], when one i is taken to right hand side to derive value of $\dfrac{-1}{i}$, the value becomes:
$i=\dfrac{-1}{i}$ . Substituting the value of $\dfrac{-1}{i}$ in expression,
Expression
  \[\begin{align}
  & =-1-\dfrac{1}{i} \\
 & =-1+i \\
\end{align}\]
The value of expression came out to be \[-1\] + i which indicates that option A is correct.

Note: All arithmetic operations – addition, subtraction, multiplication and division can be carried out on complex numbers.
Complex numbers have rules which are different from those of real numbers.
They constitute an important part of mathematical study and are mainly used to undertake research.