In the previous section, we completed a discussion on mathematical induction. In this chapter, we will see complex numbers and quadratic equations.
Let us recall the three types of equations that we have seen in our earlier classes.
(i) Linear equations in one variable
Example:
2x+5 = 17
• We know how to solve such equations. (Details here)
(ii) Linear equations in two variables
Example:
4x+3y = 43
3x - 2y = 11
• We know how to solve such equations. (Details here)
(iii) Quadratic equations in one variable
Example: x2 + 2x – 224 = 0
• We know how to solve such equations. (Details here)
• Let us consider the quadratic equations again.
We know that, the solutions of a quadratic equation can be found out by using the equation: $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Consider the portion $b^2 - 4ac$.
If this portion is -ve, we will not be able to calculate $\sqrt{b^2 - 4ac}$.
Because, root of -ve numbers do not exist.
(Mark any point on the number line. Even if that number is -ve, it’s square will be a +ve number. That is why we say that, root of a -ve number is not a real number)
• So in this chapter we try an alternate method to solve quadratic equations, when $b^2 - 4ac$ is -ve.
Significance of i
Some basics about $i$ can be written in 6 steps:
1. Consider the quadratic equation: $x^2 + 1 = 0$
• Rearranging it, we get: $x^2 = -1$
• So $x = \pm \sqrt{-1}$
2. We know that $\sqrt{-1}$ is not a real number.
• Let us denote $\sqrt{-1}$ by the symbol $i$
• Then we get: $x = \pm i$
• That is: x = $i$ or $-i$
(i) Let us substitute $i$ in the place of x in the given equation. We get: $i^2 +1 = 0$
♦ That is., $(\sqrt{-1})^2 + 1 = 0 ~\Rightarrow~ -1 + 1 = 0$, which is true.
♦ So we can write: $i$ is a solution of the equation $x^2 + 1 = 0$
(ii) Let us substitute $-i$ in the place of x in the given equation. We get: $(-i)^2 +1 = 0$
♦ That is., $(-\sqrt{-1})^2 + 1 = 0 ~\Rightarrow~ [(-1 × -1) × -1] + 1 = 0$
♦ That is.,$[(1) × -1]+1 = 0~\Rightarrow~ -1+1 =0$, which is true.
♦ So we can write: $-i$ is also a solution of the equation $x^2 + 1 = 0$
(iii) Thus we get:
♦ Solutions of the equation $x^2 + 1 = 0$ are $\sqrt{-1}$ and $-\sqrt{-1}$
♦ In other words:
♦ Solutions of the equation $x^2 + 1 = 0$ are $i$ and $-i$
| • While solving the equation $x^2 + 1 = 0$, we are actually calculating
the square roots of -1. This is clear from the fact that, the equation
is rearranged into the form: $x^2 = -1$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-1}$ can be $i$ or $-i$ • However, when we write $\sqrt{-1}$, we would mean $i$ only. ♦ If we want $-i$, we must specifically write $-\sqrt{-1}$ |
3. We denoted $\sqrt{-1}$ by the symbol $i$. What about $\sqrt{-2}$ ?
(i) Consider the quadratic equation: $x^2 + 2 = 0$
• Rearranging it, we get: $x^2 = -2$
• So $x = \pm \sqrt{-2}$
(ii) We can write $\pm \sqrt{-2}$ as:
$\pm \sqrt{-1 \times 2}~=~\pm \sqrt{-1} \times \sqrt{2}~=~ \pm i \times \sqrt{2}$
• So we get: $\sqrt{-2}~=~\pm \sqrt{2}\,i$
(iii) Let us substitute $\sqrt{2}\,i$ in the place of x in the given equation. We get: $(\sqrt{2}\,i)^2 +2 = 0$
♦ That is., $[(\sqrt{2})^2~ × ~i^2] + 2 = 0$
♦ That is., $[(\sqrt{2})^2~ × ~(\sqrt{-1})^2] + 2 = 0 ~\Rightarrow~ [2 × -1] + 2 = 0$, which is true.
♦ So we can write: $\sqrt{2}\;i$ is a solution of the equation $x^2 + 2 = 0$
(iv) Let us substitute $-\sqrt{2}\,i$ in the place of x in the given equation. We get:
$(-\sqrt{2}\,i)^2 +2 = 0$
♦ That is., $[(-1 × -1) × ~ (\sqrt{2})^2~ × ~i^2] + 2 = 0$
♦ That is., $[(-1 × -1) × ~ (\sqrt{2})^2 × ~(\sqrt{-1})^2] + 2 = 0$
♦ That is., $[(1) × 2 × -1] + 2 = 0~\Rightarrow~[-2]+2=0$, which is true.
♦ So we can write: $-\sqrt{2}\;i$ is also a solution of the equation $x^2 + 2 = 0$
(v) Thus we get:
♦ Solutions of the equation $x^2 + 2 = 0$ are $\sqrt{-2}$ and $-\sqrt{-2}$
♦ In other words:
♦ Solutions of the equation $x^2 + 2 = 0$ are $\sqrt{2}\;i$ and $-\sqrt{2}\;i$
| • While solving the equation $x^2 + 2 = 0$, we are actually calculating
the square roots of -2. This is clear from the fact that, the equation
is rearranged into the form: $x^2 = -2$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-2}$ can be $\sqrt{2}\,i$ or $-\sqrt{2}\,i$ • However, when we write $\sqrt{-2}$, we would mean $\sqrt{2}\,i$ only. ♦ If we want $-\sqrt{2}\,i$, we must specifically write $-\sqrt{-2}$ |
4. We denoted $\sqrt{-1}$ by the symbol $i$. What about $\sqrt{-3}$ ?
(i) Consider the quadratic equation: $x^2 + 3 = 0$
• Rearranging it, we get: $x^2 = -3$
• So $x = \pm \sqrt{-3}$
(ii) We can write $\sqrt{-3}$ as:
$\pm \sqrt{-1 \times 3}~=~\pm \sqrt{-1} \times \sqrt{3}~=~ \pm i \times \sqrt{3}$
• So we get: $\sqrt{-3}~=~\pm \sqrt{3} \,i$
(iii) Let us substitute $\sqrt{3}\,i$ in the place of x in the given equation. We get: $(\sqrt{3}\;i)^2 +3 = 0$
♦ That is., $[(\sqrt{3})^2~ × ~i^2] + 3 = 0$
♦ That is., $[(\sqrt{3})^2~ × ~(\sqrt{-1})^2] + 3 = 0 ~\Rightarrow~ [3 × -1] + 3 = 0$, which is true.
♦ So we can write: $\sqrt{3}\,i$ is a solution of the equation $x^2 + 3 = 0$
(iv) Let us substitute $-\sqrt{3}\,i$ in the place of x in the given equation. We get:
$(-\sqrt{3}\,i)^2 +3 = 0$
♦ That is., $[(-1 × -1) × ~ (\sqrt{3})^2~ × ~i^2] + 3 = 0$
♦ That is., $[(-1 × -1) × (\sqrt{3})^2 × (\sqrt{-1})^2] + 3 = 0$
♦ That is., $[(1) × 3 × -1] + 3 = 0~\Rightarrow~[-3]+3=0$, which is true.
♦ So we can write: $-\sqrt{3}\,i$ is also a solution of the equation $x^2 + 3 = 0$
(v) Thus we get:
♦ Solutions of the equation $x^2 + 3 = 0$ are $\sqrt{-3}$ and $-\sqrt{-3}$
♦ In other words:
♦ Solutions of the equation $x^2 + 3 = 0$ are $\sqrt{3}\,i$ and $-\sqrt{3}\,i$
| • While solving the equation $x^2 + 3 = 0$, we are actually calculating
the square roots of -3. This is clear from the fact that, the equation
is rearranged into the form: $x^2 = -3$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-3}$ can be $\sqrt{3}\,i$ or $-\sqrt{3}\,i$ • However, when we write $\sqrt{-3}$, we would mean $\sqrt{3}\,i$ only. ♦ If we want $-\sqrt{3}\,i$, we must specifically write $-\sqrt{-3}$ |
5. We denoted $\sqrt{-1}$ by the symbol $i$. What about $\sqrt{-4}$ ?
(i) Consider the quadratic equation: $x^2 + 4 = 0$
• Rearranging it, we get: $x^2 = -4$
• So $x = \pm \sqrt{-4}$
(ii) We can write $\pm \sqrt{-4}$ as:
$\pm \sqrt{-1 \times 4}~=~\pm \sqrt{-1} \times \sqrt{4}~=~ \pm i \times \sqrt{4}$
• So we get: $\sqrt{-4}~=~\pm \sqrt{4}\,i~=~\pm 2i$
(iii) Let us substitute $2i$ in the place of x in the given equation. We get: $(2i)^2 +4 = 0$
♦ That is., $[(2)^2~ × ~i^2] + 4 = 0$
♦ That is., $[(2)^2~ × ~(\sqrt{-1})^2] + 4 = 0 ~\Rightarrow~ [4 × -1] + 4 = 0$, which is true.
♦ So we can write: $2i$ is a solution of the equation $x^2 + 4 = 0$
(iv) Let us substitute $-2i$ in the place of x in the given equation. We get:
$(-2i)^2 +4 = 0$
♦ That is., $[(-1 × -1) × ~ (2)^2~ × ~i^2] + 4 = 0$
♦ That is., $[(-1 × -1) × (2)^2~ × ~(\sqrt{-1})^2] + 4 = 0$
♦ That is., $[(1) × 4 × -1] + 4 = 0~\Rightarrow~[-4]+4=0$, which is true.
♦ So we can write: $-2i$ is also a solution of the equation $x^2 + 4 = 0$
(v) Thus we get:
♦ Solutions of the equation $x^2 + 4 = 0$ are $\sqrt{-4}$ and $-\sqrt{-4}$
♦ In other words:
♦ Solutions of the equation $x^2 + 4 = 0$ are $2i$ and $-2i$
| • While solving the equation $x^2 + 4 = 0$, we are actually calculating
the square roots of -4. This is clear from the fact that, the equation
is rearranged into the form: $x^2 = -4$ • From the detailed steps that we wrote, we come to the conclusion that: $\sqrt{-4}$ can be $2i$ or $-2i$ • However, when we write $\sqrt{-4}$, we would mean $2i$ only. ♦ If we want $-2i$, we must specifically write $-\sqrt{-4}$ |
6. In this way, we can write the square roots of -5, -6, -7, . . . so on.
• In fact, we can write the square roots of any -ve real number.
• Generally, if '$a$' is a positive real number, $\sqrt{-a}=\sqrt{a} × \sqrt{-1}=\sqrt{a}\,i$
• So now we know the significance of $i$.
• Using $i$, we can do problems
involving the square roots of -ve real numbers. We will see those
problems in later sections.
• In the next section, we will see complex
numbers.
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