Question

How do you simplify $\left( 4+3i \right)\left( 7+i \right)$?

Answer 1

Hint: In this question, we have been given the multiplication of two complex numbers, $\left( 4+3i \right)$ and $\left( 7+i \right)$. A complex number is simplified by writing it in the standard form of $a+ib$, that is, we need to separately write the real and the imaginary parts. For this purpose, we have to multiply the two binomials $\left( 4+3i \right)$ and $\left( 7+i \right)$ by using the simple algebraic multiplication rule. On simplifying the expression obtained with the help of the identity ${{i}^{2}}=-1$, we will obtain the given complex number in the standard form of $a+ib$.

Complete step-by-step solution:
Let us write the complex number given in the question as
\[z=\left( 4+3i \right)\left( 7+i \right)\]
As can be seen above, we have the multiplication of two binomials. Multiplying the two binomials using the algebraic rule of multiplication, we get
$\begin{align}
  & \Rightarrow z=4\left( 7+i \right)+3i\left( 7+i \right) \\
 & \Rightarrow z=4\left( 7 \right)+4\left( i \right)+3i\left( 7 \right)+3i\left( i \right) \\
 & \Rightarrow z=28+4i+21i+3{{i}^{2}} \\
 & \Rightarrow z=28+25i+3{{i}^{2}}........(i) \\
\end{align}$
Now, we know that the complex number $i$ is defined as the square root of negative unity. So we can write
$i=\sqrt{-1}$
On taking the squares of both the sides, we get
$\begin{align}
  & \Rightarrow {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}} \\
 & \Rightarrow {{i}^{2}}={{\left( -1 \right)}^{\dfrac{1}{2}\times 2}} \\
 & \Rightarrow {{i}^{2}}={{\left( -1 \right)}^{1}} \\
 & \Rightarrow {{i}^{2}}=-1........(ii) \\
\end{align}$
Substituting the equation (ii) in the equation (i), we get
$\begin{align}
  & \Rightarrow z=28+25i+3\left( -1 \right) \\
 & \Rightarrow z=28+25i-3 \\
 & \Rightarrow z=28-3+25i \\
 & \Rightarrow z=25+25i \\
\end{align}$
Finally, on taking $25$ common from both the terms, we get the simplified form of the given complex number as
$\Rightarrow z=25\left( 1+i \right)$
Hence, the complex number $\left( 4+3i \right)\left( 7+i \right)$ is simplified as $25\left( 1+i \right)$.

Note: Do not take ${{i}^{2}}=1$. This is a common confusion to write ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=\sqrt{{{\left( -1 \right)}^{2}}}=1$. This is not the right way to calculate the value of ${{i}^{2}}$. The order of the square and the square root cannot be changed. Also, the equation ${{i}^{2}}=-1$ must be remembered as an identity.