Question

Let \[z\] be the complex number $-1+\sqrt{3}i$.
a) Express ${{z}^{2}}$ in the form of $a+ib$.

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 solution:
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 $2xy$. We will use this property to solve this problem.

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,
$\Rightarrow z=-1+\sqrt{3}i........(iii)$
Now, ${{z}^{2}}$ is represented as,
$\begin{align}
  & \Rightarrow {{z}^{2}}=\left( -1+i\sqrt{3} \right)\left( -1+i\sqrt{3} \right) \\
 & \Rightarrow {{z}^{2}}=\left( {{\left( -1 \right)}^{2}}-i\sqrt{3}-i\sqrt{3}+{{i}^{2}}{{\left( \sqrt{3} \right)}^{2}} \right) \\
 & \Rightarrow {{z}^{2}}=(1-i2\sqrt{3}+3{{i}^{2}})..........(iv) \\
\end{align}$
In equation (iv) we have ${{i}^{2}}$. From equation (ii) we know that ${{i}^{2}}=-1$. Substituting this in equation (iv) we get equation (iv) as,
$\Rightarrow {{z}^{2}}=(1-i2\sqrt{3}-3)..........(v)$
Now, again simplifying equation (v) we get equation (v) as,
$\Rightarrow {{z}^{2}}=(-2-i2\sqrt{3})..........(vi)$
Here, ${{z}^{2}}$ is expressed as $a+ib$. Where, the real part ,$a=-2$ and the imaginary part, $b=-2\sqrt{3}$.
Hence, if \[z\] is the complex number $-1+\sqrt{3}i$, then ${{z}^{2}}$ expressed in the form of $a+ib$ is $-2-i2\sqrt{3}$.
Therefore, the correct answer is ${{z}^{2}}=-2-i2\sqrt{3}$.

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, $z=-1+\sqrt{3}i$
Directly substituting this we get,
$\begin{align}
  & \Rightarrow {{z}^{2}}=\left( {{x}^{2}}-{{y}^{2}} \right)+i2xy \\
 & \Rightarrow {{z}^{2}}=\left( {{\left( -1 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)+i\left( 2\times -1\times \sqrt{3} \right) \\
 & \Rightarrow {{z}^{2}}=(1-3)+i\left( -2\sqrt{3} \right) \\
 & \Rightarrow {{z}^{2}}=-2-2\sqrt{3}i. \\
\end{align}$