Question

 Evaluate the following, expressing your answer in Cartesian form $a+ib$:
(a) $\left( 1+2i \right){{(4-6i)}^{2}}$

Answer 1

Hint: Various properties of complex numbers are used to solve this problem. The properties used in this problem are:
\[\begin{align}
  & \Rightarrow \sqrt{-1}=i \\
 & \Rightarrow {{i}^{2}}=-1 \\
\end{align}\]
Complete step-by-step answer:
A complex number $z$ has two parts. A real part and an imaginary part. If $z=x+iy$ is a complex number. Then ${{z}^{2}}$ is represented as,
$\begin{align}
  & \Rightarrow {{z}^{2}}=(x+iy)(x+iy) \\
 & \Rightarrow {{z}^{2}}=({{x}^{2}}+ixy+ixy+{{i}^{2}}{{y}^{2}}) \\
 & \Rightarrow {{z}^{2}}=({{x}^{2}}-{{y}^{2}}+i2xy) \\
 & \Rightarrow {{z}^{2}}={{x}^{2}}-{{y}^{2}}+i2xy \\
\end{align}$
This means that the real part of ${{z}^{2}}$ is ${{x}^{2}}-{{y}^{2}}$ and the imaginary part of ${{z}^{2}}$ is $2xy$.
 If ${{z}_{1}}=a+ib$and ${{z}_{2}}=c+id$ are two complex numbers then multiplication of ${{z}_{1}}$ and ${{z}_{2}}$ is denoted by ${{z}_{1}}{{z}_{2}}$.
$\begin{align}
  & \Rightarrow {{z}_{1}}{{z}_{2}}=(a+ib)(c+id) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=(ac+iad+ibc+{{i}^{2}}bd) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=(ac-bd+iad+ibc) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=ac-bd+i(ad+bc) \\
\end{align}$
This means that the real part of ${{z}_{1}}{{z}_{2}}$ is $ac-bd$ and the imaginary part of ${{z}_{1}}{{z}_{2}}$ is $ad+cb$. We will be using the above properties of complex numbers to find the answer.

The symbol $i$(iota) is used to represent the square root of $\sqrt{-1}$. This also implies, ${{i}^{2}}=-1$, which means $i$ is the solution of the quadratic equation ${{x}^{2}}+1=0$. In this way the square root of any negative number can be expressed using $i$.
$\begin{align}
  & \Rightarrow \sqrt{-1}=i.......(i) \\
 & \Rightarrow {{i}^{2}}=-1........(ii) \\
\end{align}$
Here, we have the expression,
$\Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}$.
First, we will evaluate ${{\left( 4-6i \right)}^{2}}$. For this we have,
\[\begin{align}
  & \Rightarrow {{\left( 4-6i \right)}^{2}}=\left( 4-6i \right)\left( 4-6i \right) \\
 & \Rightarrow {{\left( 4-6i \right)}^{2}}=\left( {{4}^{2}}-2\times 4\times 6i+{{\left( 6i \right)}^{2}} \right) \\
 & \Rightarrow {{\left( 4-6i \right)}^{2}}=\left( 16-48i+36{{i}^{2}} \right)..........(iii) \\
\end{align}\]
In equation (iii) we have ${{i}^{2}}$, and we know ${{i}^{2}}=-1$ from equation (ii). Substituting this in equation (iii) we get,
$\begin{align}
  & \Rightarrow {{\left( 4-6i \right)}^{2}}=\left( 16-36-48i \right) \\
 & \Rightarrow {{\left( 4-6i \right)}^{2}}=\left( -20-48i \right)..........(iv) \\
\end{align}$
Now, substituting equation (iv) in the expression $\left( 1+2i \right){{\left( 4-6i \right)}^{2}}$we get,
$\Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}=\left( 1+2i \right)\left( -20-48i \right)........(v)$
 Equation (v) can be again simplified as,
$\begin{align}
  & \Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}=4\left( 1+2i \right)\left( -5-12i \right) \\
 & \Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}=4\left( -5-12i-10i-24{{i}^{2}} \right)......(vi) \\
\end{align}$
In equation (vi) we have ${{i}^{2}}$, and we know ${{i}^{2}}=-1$ from equation (ii). Substituting this in equation (vi) we get,
\[\begin{align}
  & \Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}=4\left( -5-12i-10i+24 \right) \\
 & \Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}=4\left( 19-22i \right) \\
 & \Rightarrow \left( 1+2i \right){{\left( 4-6i \right)}^{2}}=76-88i..........(vii) \\
\end{align}\]
From equation (vii) we get the cartesian form of the expression $\left( 1+2i \right){{\left( 4-6i \right)}^{2}}$.
Hence, the correct answer is $76-88i$.

Note: In this problem we can also find the value ${{z}^{2}}$using the direct formula${{z}^{2}}=\left( {{x}^{2}}-{{y}^{2}} \right)+i2xy$.
Where, $z=x+iy$. In this question, \[z=4-6i\]
Directly substituting this we get,
$\begin{align}
  & \Rightarrow {{z}^{2}}=\left( {{x}^{2}}-{{y}^{2}} \right)+i2xy \\
 & \Rightarrow {{z}^{2}}=\left( {{\left( 4 \right)}^{2}}-{{\left( -6 \right)}^{2}} \right)+i\left( 2\times -6\times 4 \right) \\
 & \Rightarrow {{z}^{2}}=(16-36)+i\left( -48 \right) \\
 & \Rightarrow {{z}^{2}}=-20-48i. \\
\end{align}$
And after this $\left( 1+2i \right){{\left( 4-6i \right)}^{2}}$can be also found using the direct formula,$pq=(ac-bd)+i(ad+bc)$.
Here,\[p=1+2i\],this implies \[a=1,b=2\]. Similarly, \[q=-20-48i\] implies \[c=-20,d=-48\].
Directly substituting this we get,
\[\begin{align}
  & \Rightarrow pq=(ac-bd)+i(ad+bc) \\
 & \Rightarrow pq=\left( \left( 1\times -20 \right)-\left( 2\times -48 \right) \right)+i\left( \left( 1\times -48 \right)+\left( 2\times -20 \right) \right) \\
 & \Rightarrow pq=\left( -20+96 \right)+i\left( -48-40 \right) \\
 & \Rightarrow pq=76-88i. \\
\end{align}\]