Question

If $i=\sqrt{-1}$, then calculate the value of \[(3+i)(3-i)\].
A. 8
B. 9
C. 10
D. 9 – i
E. 9 + i

Answer 1

Hint: We know the value of complex number i is $\sqrt{-1}$. We generally represent it by i and therefore ${{i}^{2}}=-1$. Hence, we can expand the expression by multiplying both the terms in the brackets and then substitute the known values.

Complete step-by-step answer:
The value of i is $\sqrt{-1}$. Therefore ${{i}^{2}}=-1$.

Given expression is \[(3+i)(3-i)\]
On multiplying the terms in the expression,

\[\Rightarrow (3+i)(3-i)=\,3\times 3+3i-3i-{{i}^{2}}\]
In the above equation \[+3i-3i\] gets canceled and becomes 0.
\[\Rightarrow (3+i)(3-i)=\,9-{{i}^{2}}\]

Putting the value of ${{i}^{2}}=-1$ in the above equation, the equation $\left( 3+i \right)\left( 3-i \right)=9-{{i}^{2}}$ becomes:

\[\Rightarrow (3+i)(3-i)=\,9-(-1)\]
\[\Rightarrow (3+i)(3-i)=\,9+1\]
\[\Rightarrow (3+i)(3-i)=\,10\]

Therefore the value \[(3+i)(3-i)\] is equal to 10.
Hence option C. 10 is the correct answer.

Note: While solving questions we have to take care about negative signs because negative times negative equal to positive. This problem can be solved using another direct method that is, by using the identity $(a-b)(a+b) = a^2-b^2$, where a = 3 and b = i.