Question

How do you multiply ${(3 + 2i)^2}$ ?

Answer 1

As we can see the question is in the form of ${(a + b)^2}$, here we will apply the identity ${(a +b)^2} = {a^2} + 2ab + {b^2}$ to open the parentheses and then we will simplify the terms obtained individually. Since the term is a complex number which contains an imaginary term iota i.e., $i$, another concept that can be used in this question is ${i^2} = - 1$.

Complete step by step solution:
As we know the identity,
${(a + b)^2} = {a^2} + 2ab + {b^2}$
So, applying this identity in our question, it becomes:
${(3 + 2i)^2} = {(3)^2} + 2(3)(2i) + {(2i)^2}$
On simplifying each term individually, we get:
${(3 + 2i)^2} = 9 + 12i + 4{i^2}$ -(eq.1)
As we know that the value of iota $i$ is $\sqrt { - 1}$ i.e.,
$i = \sqrt { - 1}$
When we square both the sides, it becomes:
${i^2} = {(\sqrt { - 1} )^2}$
As we know that the square of the root of any number is the number itself i.e., ${(\sqrt x )^2} = x$. In
simpler words, here the square will get cut with the root and the result would be $- 1$.
Therefore, ${i^2} = - 1$
Putting the value of ${i^2}$as $- 1$ in eq.1, we get:
${(3 + 2i)^2} = 9 + 12i + 4( - 1)$
Opening the parentheses and multiplying $4$ with $- 1$, it will become:
${(3 + 2i)^2} = 9 + 12i - 4$
As we can add real numbers with each other and imaginary numbers with each other, here the two terms of real numbers $9$ and $- 4$ can be added together.
So, we obtain: ${(3 + 2i)^2} = 5 + 12i$
Hence, multiplying ${(3 + 2i)^2}$, the answer obtained is $5 + 12i$.

Note: One of the alternate ways to solve this question could be opening the parentheses directly by multiplying $(3 + 2i)$ by itself. We will obtain the same answer even without using the identity, by simple multiplication, addition and subtraction. The most important step here is opening parentheses either by applying the identity or by multiplying it directly so do it carefully.