Simplify\[\left( {3 + i} \right)\left( {3 + i} \right)\].

Hint: We know that:
 $
  \left( {a + b} \right)\left( {a + b} \right) = {a^2} + ab + ab + {b^2} \\
   = {a^2} + 2ab + {b^2} \;
  $
On comparing the question and the above identity we see that they are similar except the second term $ b $ is an imaginary term in the question whereas it’s real in the identity. So we can solve the question using the above identity and some basic properties of imaginary numbers that are given below:
 $
  i = \sqrt { - 1} \\
  {i^2} = - 1\; \;
  $
So by using the above information we can solve this question.

Complete step-by-step answer:
Given
\[\left( {3 + i} \right)\left( {3 + i} \right).....................\left( i \right)\]
Now we also know
\[
  \left( {a + b} \right)\left( {a + b} \right) = {a^2} + ab + ab + {b^2} = {a^2} + 2ab + {b^2}........................\left( {ii} \right) \;
 \]
Now applying (ii) in (i) we get:
\[\left( {3 + i} \right)\left( {3 + i} \right) = \left( {3 \times 3} \right) + 3i + 3i + \left( {i \times i} \right)...........\left( {iii} \right)\]
\[ \Rightarrow \left( {3 \times 3} \right) + 3i + 3i + \left( {i \times i} \right) = 9 + 6i + {i^2}.............\left( {iv} \right)\]
Now in (iv) we can further simplify $ {i^2} $ using the basic properties of imaginary numbers which is:
\[
  i = \sqrt { - 1} \\
  {i^2} = \left( {\sqrt { - 1} \times \sqrt { - 1} } \right) = - 1\;
 \]
So substituting the value of $ {i^2} $ in (iv), we get:
 $
 \Rightarrow 9 + 6i + {i^2} = 9 + 6i - 1 \\
 = 8 + 6i.................\left( v \right) \;
  $
So on observing (v) we can see that it cannot be simplified any further, such that we can stop the process of simplification in the (v) step.
Therefore on simplifying \[\left( {3 + i} \right)\left( {3 + i} \right)\] we get $ 8 + 6i. $
Such that $ 8 + 6i $ is our final answer.
So, the correct answer is “ $ 8 + 6i $ ”.

Note:
Any other powers of $ i $ higher or lower than the above mentioned ones can be expressed as multiples of any of the above given values.
Another property widely popular and used for solving imaginary powers are the exponential properties, which form the basis for solving many problems with imaginary numbers.
Also $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $ : It can be used if the question is expressed in $ \left( {a - b} \right)\left( {a - b} \right) $ format.
Complex numbers are the numbers expressed in the form of $ a + bi $ where ‘a’ is the real part and ‘b’ is the imaginary part is the field where imaginary numbers are important and are of great use.
Some useful formulas for solving questions containing imaginary numbers:
 $
  i = \sqrt { - 1} \\
  {i^2} = - 1\; \\
  {i^3} = - i \\
  {i^4} = 1 \\
  {i^{4n\;}} = 1 \\
  {i^{4n - 1}} = - i \;
  $