Question

How do you simplify ${\left( {2 + 3i} \right)^2}$?

Answer 1

Here, we will simply expand the expression given to us by using a suitable algebraic identity. Then we will simplify the iota terms by using the value of iota. Finally, we will calculate the values to get our desired answer.

Formula used:
${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$

Complete step by step solution:
We have to simplify ${\left( {2 + 3i} \right)^2}$
Now, we will first expand the given expression.
Using the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$, we get
${\left( {2 + 3i} \right)^2} = {2^2} + 2 \times 2 \times 3i + {\left( {3i} \right)^2}$
Simplifying the expression, we get
$\Rightarrow {\left( {2 + 3i} \right)^2} = 4 + 12i + 9{i^2}$……$\left( 1 \right)$
We know that $i = \sqrt { - 1}$ and ${i^2} = - 1$.
So, substituting value of ${i^2} = - 1$ in equation $\left( 1 \right)$, we get
${\left( {2 + 3i} \right)^2} = 4 + 12i + 9 \times \left( { - 1} \right)$
Multiplying the terms, we get
$\Rightarrow {\left( {2 + 3i} \right)^2} = 4 + 12i - 9$
Adding and subtracting the like terms, we get
$\Rightarrow {\left( {2 + 3i} \right)^2} = - 5 + 12i$

Therefore, the value of ${\left( {2 + 3i} \right)^2}$is $- 5 + 12i$.

Additional information:
The numbers which don’t fall anywhere in the number line are known as imaginary numbers also known as complex numbers. The square root of negative numbers is termed Complex number where we denote them in the form of iota. All the numbers having iota i.e. $i$ in them are known as imaginary numbers. Imaginary numbers are an important concept of mathematics that extends the real number system to the complex number system. We can find the cube root and square root of the complex numbers. Multiplication and division can also be done on complex numbers.

Note:
Algebraic Identities are algebraic equations that are always true for any given value of the variables. The value on the right side of the identity is always equal to values on the left side of the equation. When a number is multiplied by itself it is called the square of the number. Here, we have solved the square of a complex number and we got the answer in complex form only such that $- 5$ is the real part and $12i$ is the imaginary part.