1. Imaginary and Complex Numbers
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In this blog, we will discuss and understand:
- ✓ What imaginary numbers and complex numbers are
- ✓ How to add and subtract complex numbers
- ✓ How to multiply complex numbers
- ✓ How to divide complex numbers using complex conjugates
Introduction
As we have all dealt with quadratic equations in our GCSEs, $ax^2 + bx + c = 0$, its solutions are given by the quadratic formula:
In your GCSEs, whenever the term under the square root, $b^2 – 4ac$, was negative, you were told that there were no real solutions.
What are Imaginary Numbers?
Roots can actually be found even if the square root term is negative, and such roots are known as imaginary numbers. You might have even mistakenly stumbled across the symbol $i$ on your calculator when getting an unexpected answer.
What you need to remember here is that the square root of a negative number gives imaginary numbers, where $i = \sqrt{-1}$.
1. $\sqrt{-9}$:
$$\sqrt{-9} = \sqrt{9 \times (-1)} = \sqrt{9} \times \sqrt{-1} = 3i$$
2. $\sqrt{-32}$:
$$\sqrt{-32} = \sqrt{32 \times (-1)} = \sqrt{16} \times \sqrt{2} \times \sqrt{-1} = 4\sqrt{2}i$$
Numbers of the form $bi$, where $b \in \mathbb{R}$, are known as imaginary numbers.
What are Complex Numbers?
Just as we have imaginary numbers, we also have complex numbers.
A complex number is made up of two parts: a real part and an imaginary part.
For example, in the number $3 + 2i$:
- $3$ is the real part
- $2i$ is the imaginary part
In standard mathematical notation, a complex number is represented by the letter $z$:
For the complex number $z = a + bi$:
- $\text{Re}(z) = a$ is the real part
- $\text{Im}(z) = b$ is the imaginary part
(Note: The set of all complex numbers is represented by the symbol $\mathbb{C}$.)
2. Addition and Subtraction of Complex Numbers
Complex numbers can be added and subtracted in a very similar way to standard algebraic expressions—by gathering like terms! You simply add or subtract their real parts together, and add or subtract their imaginary parts together.
$$(2 + 6i) + (3 + 4i) = (2 + 3) + (6 + 4)i = 5 + 10i$$
$$(2 – 5i) – (5 – 11i) = (2 – 5) + (-5 – (-11))i = -3 + 6i$$
Scalar Multiplication and Division
You can also multiply or divide a complex number by a real constant just as you would expand or simplify brackets in algebra.
$$2(5 – 8i) = (2 \times 5) – (2 \times 8)i = 10 – 16i$$
$$\frac{10 + 6i}{2} = \frac{10}{2} + \frac{6}{2}i = 5 + 3i$$
Group the real parts and imaginary parts separately:
$$\text{Real part: } 18 – 15 – 3 = 0$$
$$\text{Imaginary part: } (5 – (-2) – 7)i = (5 + 2 – 7)i = 0i$$
Final Answer: $0$
a + bi using exact fractions where needed.Separate each fraction into real and imaginary parts:
$$\left(-\frac{8}{4} + \frac{3}{4}i\right) – \left(\frac{7}{2} – \frac{2}{2}i\right) = \left(-2 + \frac{3}{4}i\right) – \left(\frac{7}{2} – i\right)$$
Group the real and imaginary parts:
$$\text{Real part: } -2 – \frac{7}{2} = -\frac{4}{2} – \frac{7}{2} = -\frac{11}{2}$$
$$\text{Imaginary part: } \frac{3}{4}i – (-i) = \left(\frac{3}{4} + 1\right)i = \frac{7}{4}i$$
Final Answer: $-\frac{11}{2} + \frac{7}{4}i$
3. Solving Quadratic Equations with Complex Roots
Now that we know what imaginary numbers are, we can use complex numbers to find the solutions to any quadratic equation with real coefficients, even when $b^2 – 4ac < 0$.
Using $z = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$ with $a = 1$, $b = 6$, and $c = 25$:
$$z = \frac{-6 \pm \sqrt{6^2 – 4(1)(25)}}{2(1)}$$
$$z = \frac{-6 \pm \sqrt{36 – 100}}{2} = \frac{-6 \pm \sqrt{-64}}{2}$$
Since $\sqrt{-64} = \sqrt{64}\sqrt{-1} = 8i$:
$$z = \frac{-6 \pm 8i}{2} = -3 \pm 4i$$
So, the two complex solutions are $z_1 = -3 + 4i$ and $z_2 = -3 – 4i$.
Apply the quadratic formula with $a = 7$, $b = -3$, $c = 3$:
$$z = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(7)(3)}}{2(7)} = \frac{3 \pm \sqrt{9 – 84}}{14} = \frac{3 \pm \sqrt{-75}}{14}$$
Simplify $\sqrt{-75} = \sqrt{25 \times 3 \times (-1)} = 5\sqrt{3}i$:
$$z = \frac{3 \pm 5\sqrt{3}i}{14} = \frac{3}{14} \pm \frac{5\sqrt{3}}{14}i$$
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You can multiply complex numbers using the exact same expansion techniques as expanding two brackets in algebra, such as the FOIL method:
The key thing to remember when expanding complex numbers is how powers of $i$ behave:
- $i = \sqrt{-1}$
- $i^2 = -1$
- $i^3 = i^2 \times i = (-1) \times i = -i$
- $i^4 = i^2 \times i^2 = (-1) \times (-1) = 1$
Expanding the brackets:
$$(2 – 3i)(4 – 5i) = 2(4 – 5i) – 3i(4 – 5i)$$
$$= 8 – 10i – 12i + 15i^2$$
Since $i^2 = -1$, substitute $15i^2 = 15(-1) = -15$:
$$= 8 – 22i – 15 = (8 – 15) – 22i = -7 – 22i$$
Final Answer: $-7 – 22i$
Expand the squared term:
$$(2 – 9i)^2 = (2 – 9i)(2 – 9i) = 4 – 18i – 18i + 81i^2$$
Substitute $i^2 = -1$:
$$= 4 – 36i – 81 = -77 – 36i$$
Substitute $z = 1 – 4i$ into $f(z)$:
$$f(1 – 4i) = (1 – 4i)^2 – 2(1 – 4i) + 17$$
Expand $(1 – 4i)^2 = 1 – 8i + 16i^2 = -15 – 8i$.
Expand $-2(1 – 4i) = -2 + 8i$.
Combine terms:
$$f(1 – 4i) = (-15 – 8i) + (-2 + 8i) + 17 = (-15 – 2 + 17) + (-8i + 8i) = 0$$
Since $f(1 – 4i) = 0$, $z = 1 – 4i$ is a solution to $f(z) = 0$. (The other root is $1 + 4i$).
5. Complex Conjugates and Division of Complex Numbers
What is a Complex Conjugate?
For any complex number $z = a + bi$, its complex conjugate is denoted by $z^*$ and is defined as:
To find the complex conjugate, all you need to do is change the sign of the imaginary part (flip $+ \to -$ or $- \to +$). The real part stays exactly the same!
Examples:
- If $z = 2 + 3i$, then its complex conjugate is $z^* = 2 – 3i$.
- If $z = 3 – 4i$, then its complex conjugate is $z^* = 3 + 4i$.
A special property of complex conjugates is that when you multiply a complex number $z$ by its conjugate $z^*$, the result is always a real number:
$$(a + bi)(a – bi) = a^2 – abi + abi – b^2 i^2 = a^2 + b^2$$
Dividing Complex Numbers
Dividing complex numbers is very similar to rationalising the denominator when working with surds! Just as we cannot leave a surd (like $\sqrt{2}$) in the denominator of a fraction, we cannot leave an imaginary number ($i$) in the denominator.
The denominator is $4 – 2i$, so its complex conjugate is $4 + 2i$.
$$\frac{7 + 3i}{4 – 2i} = \frac{7 + 3i}{4 – 2i} \times \frac{4 + 2i}{4 + 2i} = \frac{(7 + 3i)(4 + 2i)}{(4 – 2i)(4 + 2i)}$$
Numerator: $(7 + 3i)(4 + 2i) = 28 + 14i + 12i + 6i^2 = 22 + 26i$
Denominator: $(4 – 2i)(4 + 2i) = 16 + 8i – 8i – 4i^2 = 20$
$$\frac{22 + 26i}{20} = \frac{22}{20} + \frac{26}{20}i = \frac{11}{10} + \frac{13}{10}i$$
Final Answer: $\frac{11}{10} + \frac{13}{10}i$
Multiply numerator and denominator by conjugate $2 + i$:
$$\frac{(28 – 3i)(2 + i)}{(2 – i)(2 + i)} = \frac{56 + 28i – 6i – 3i^2}{4 – i^2} = \frac{59 + 22i}{5}$$
Final Answer: $\frac{59}{5} + \frac{22}{5}i$
(ap - b)/d - ((cp + e)/d)i.Multiply numerator and denominator by conjugate $2 – 5i$:
$$\frac{(p – 7i)(2 – 5i)}{(2 + 5i)(2 – 5i)} = \frac{2p – 5pi – 14i + 35i^2}{4 + 25}$$
$$= \frac{(2p – 35) – (5p + 14)i}{29}$$
Final Answer: $\frac{2p – 35}{29} – \left(\frac{5p + 14}{29}\right)i$
