Question

How do you multiply \[\left( 3+2i \right)\left( 3-2i \right)\] ?

Answer 1

Hint: These types of problems are very straight forward and are simple to solve. Anything which is written of the form \[a+ib\] represents a complex number, where ‘a’ is the real part and ‘b’ is the imaginary part. In this problem of multiplication of two complex numbers, it is done by using normal algebraic multiplication, only the addition part is done by adding only the like terms together, that is, adding only the real parts together and the imaginary parts together. We know \[i\] is basically known as aorta or the imaginary part which is defined as,
\[i=\sqrt{-1}\] and \[{{i}^{2}}=-1\]

Complete step by step answer:
Now proceeding with the solution, we do simple algebraic multiplication to get,
\[\begin{align}
  & \left( 3+2i \right)\left( 3-2i \right) \\
 & =\left( 3\times 3 \right)+\left( 2i\times 3 \right)+\left( 3\times \left( -2i \right) \right)+\left( 2i\times \left( -2i \right) \right) \\
\end{align}\]
Now, evaluating the terms we get,
\[9+6i+\left( -6i \right)+\left( -4{{i}^{2}} \right)\]
Now, taking out the negative sign out of the equation we get,
\[9+6i-6i-4{{i}^{2}}\]
Now cancelling the \[+6i\] and \[-6i\], we get,
\[9-4{{i}^{2}}\]
Now, from the rules of the complex numbers, we can say that \[{{i}^{2}}=-1\] , thus putting this value in our intermediate equation, we get,
\[9-4\left( -1 \right)\]
Now, taking out the negative sign, we can further write it as,
\[9+4\]
Adding, we get,
\[13\]

Note:
The given problem can also be done in another method using the concept of factors. We can clearly see that the above given problem is a simple factor of the form \[\left( a+b \right)\left( a-b \right)\] which is a factor of \[{{a}^{2}}-{{b}^{2}}\] . Applying this theory in this problem, we can solve it very easily. We must also be careful in adding and multiplying of the complex numbers and should not mix up between the two or else solving the problem might be difficult.