Dividing Complex Numbers
Learn how to divide complex numbers with ease.
๐ Introduction
In this lesson, we will explore the process of dividing complex numbers. Understanding this concept is crucial for solving advanced problems in mathematics, particularly in fields like engineering and physics. It seems tricky, but it's quite easy.
๐ฏ Learning Objectives
By the end of this lesson, you will be able to:
- Understand the concept of dividing complex numbers.
- Apply the division algorithm to complex numbers.
- Understand what happens when we divide.
๐ Key Concepts
Must knows before tackling division
Before we dive into dividing complex numbers, it's essential to grasp a few key concepts:
What complex numbers are (in the form a + bi).
The concept of the conjugate of a complex number.
๐ Definition
Conjugate Of Complex number: If $z = a + bi, \quad a, b \in \mathbb{R}$, then the conjugate is: $z = a - bi$
Its simply just flipping the sign in a complex number.
If a complex number is $z$, then the conjugate is denoted as:
$$\overline{z}\quad \text{or,}\quad z^{-1}\quad \text{or,}\quad z^*$$
Simple division of Complex numbers
Dividing a complex number by a real number is very easy.
Just divide both the real and imaginary parts by that number.
๐ Example of Simple Division
$z = 9 + 12i$. Solve: $\frac{z}{3}$
To divide by 3, we divide both the real and imaginary parts by 3:
$$\frac{z}{3} = \frac{9 + 12i}{3} = \frac{9}{3} + \frac{12i}{3} = 3 + 4i$$
Complex Division
This is where you divide a Complex number by another complex number.
To do this, we multiply the top and bottom by the conjugate of the denominator.
Allow me to show you an example (after the tip).
๐ก Pro Tip
For simple division, its really easy if you write $a$ and $b$ clearly. Then, you can simply do: $$\frac{a}{\text{denominator}} + \frac{bi}{\text{denominator}}$$
Remember, to find the conjugate, just flip the sign. Example:
$$\text{let } z = 3 + 4i, \text{ then } \overline{z} = 3 - 4i$$
Dont forget, the conjugate is also sometimes given as:
$$z^*, \text{ or } z^{-1}$$
๐ Worked Examples
๐ Example 1: Dividing Complex number by another complex number.
Question: If $z = 5 + 10i$, and $w = 2 + 3i$, solve: $\frac{z}{w}$
Solution:
Find the conjugate of the Denominator - just flip the sign.
Now multiply the numerator and denominator by the conjugate of the denominator.
The question asked for $\frac{z}{w}$, so write it clearly:
$\text{Multiply top and bottom by the conjugate of denominator:}$ $$= \frac{(5 + 10i)(2 - 3i)}{(2 + 3i)(2 - 3i)}$$
Let's solve the denominator first.
We have: $(2 + 3i)(2 - 3i)$ so let's expand the brackets.
Remember, $i^{2} = -1, \text{ because: } i = \sqrt{-1}, \text{ and } (\sqrt{-1})^2 = -1$
Let's solve the numerator now.
We have: $(5 + 10i)(2 - 3i)$ so let's expand the brackets.
Remember, $i^{2} = -1, \text{ because: } i = \sqrt{-1}, \text{ and } (\sqrt{-1})^2 = -1$
Now we can complete the division by putting our results together:
Its best to give exact answers, so you can leave the answer in fraction form.
โ ๏ธ Common Mistakes
๐ซ Mistake 1: Forgetting $i^2 = -1$
What students often do wrong: They forget to apply the fact that $i^2 = -1$ when simplifying expressions involving $i$.
Why it's wrong: This leads to incorrect simplifications and ultimately wrong answers.
How to avoid it: Always remember to substitute $i^2$ with $-1$ when simplifying expressions.
๐ซ Mistake 2: Adding like terms incorrectly, or forgetting it completely.
What students often do wrong: They either forget to combine like terms or do it incorrectly.
Why it's wrong: This can easily lead to incorrect answers, especially in complex expressions.
How to avoid it: Always double-check your work to ensure that like terms are combined correctly, and practice further.
๐ช Practice Problems
Try these problems to test your understanding:
๐ฏ Practice Questions A
(A) $j = 3 + 4i$, and $k = 5 - 2i$, find $j^{*}$ and $k^{*}$
(B) $m = 64 + 48i$, Solve simple division: $\frac{m}{8}$
(C) $p = 8 + 6i$, and $q = 2 + 3i$, Solve: $p^{*}+q^{*}$
(D) $q = 2 + 3i$. find $qq^{*}$
๐ Show Solution
(A) $j^{*} = 3 - 4i$, and $k^{*} = \boxed{5 + 2i}$
(B) $\frac{m}{8} = \frac{64 + 48i}{8} = \frac{64}{8} + \frac{48i}{8} = \boxed{8 + 6i}$
(C) $p^{*}+q^{*} = (8 - 6i) + (2 - 3i) = \boxed{10 - 9i}$
(D) $qq^{*} = (2 + 3i)(2 - 3i) = 4 - 9i^{2} = 4 + 9 = \boxed{13}$
๐ฏ Practice Questions B
(A) $z = 7 + 6i$, and $w = 3 + 2i$, Solve: $\frac{z}{w}$
(B) $a = 4 + 5i$, and $b = 1 - 2i$, Solve: $\frac{a}{b}$
(C) $c = 1 + i$, and $d = 1 - i$, Solve: $\frac{c}{d}$
(D) $e = 2 + 3i$, and $f = 4 + i$, Solve: $\frac{e}{f}$
๐ Show Solution
(A) $\frac{z}{w} = \frac{(7 + 6i)(3 - 2i)}{(3 + 2i)(3 - 2i)} = \frac{21 - 14i + 18i - 12i^{2}}{9 - 4i^{2}} = \frac{21 + 4i + 12}{9 + 4} = \frac{33 + 4i}{13} = \boxed{\frac{33}{13} + \frac{4}{13}i}$
(B) $\frac{a}{b} = \frac{(4 + 5i)(1 + 2i)}{(1 - 2i)(1 + 2i)} = \frac{4 + 8i + 5i + 10i^{2}}{1 - 4i^{2}} = \frac{4 + 13i - 10}{1 + 4} = \frac{-6 + 13i}{5} = \boxed{-\frac{6}{5} + \frac{13}{5}i}$
(C) $\frac{c}{d} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + i + i + i^{2}}{1 - i^{2}} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = \boxed{i}$
(D) $\frac{e}{f} = \frac{(2 + 3i)(4 - i)}{(4 + i)(4 - i)} = \frac{8 - 2i + 12i - 3i^{2}}{16 - i^{2}} = \frac{8 + 10i + 3}{16 + 1} = \frac{11 + 10i}{17} = \boxed{\frac{11}{17} + \frac{10}{17}i}$
๐ Challenge Exam Question
๐ Exam-Style Question
Give your answers in exact value.
$h = \frac{6 + 8i}{2 + 3i}$, and $j = \frac{5 + 7i}{1 - 4i}$.
(A) Find the modulus of $h$ (3 marks)
(B) Find the argument of $j$. Give your answer in Radians. (3 marks)
(C) Solve $h + j$ (4 marks)
๐ Show Solution
(A) First, find $h$:
$$h = \frac{6 + 8i}{2 + 3i} = \frac{(6 + 8i)(2 - 3i)}{(2 + 3i)(2 - 3i)}$$
$$= \frac{12 - 18i + 16i - 24i^{2}}{4 - 9i^{2}} = \frac{12 - 2i + 24}{4 + 9}$$
$$= \frac{36 - 2i}{13} = \frac{36}{13} - \frac{2}{13}i$$
Now find the modulus of $h$:
$$|h| = \sqrt{\left(\frac{36}{13}\right)^2 + \left(-\frac{2}{13}\right)^2}$$
$$= \sqrt{\frac{1296}{169} + \frac{4}{169}} = \sqrt{\frac{1300}{169}}$$
$$= \frac{\sqrt{1300}}{13} = \frac{\sqrt{100 \cdot 13}}{13} = \frac{10\sqrt{13}}{13}$$
$$\boxed{|h| = \frac{10\sqrt{13}}{13}}$$
(B) First, find $j$:
$$j = \frac{5 + 7i}{1 - 4i} = \frac{(5 + 7i)(1 + 4i)}{(1 - 4i)(1 + 4i)}$$
$$= \frac{5 + 20i + 7i + 28i^{2}}{1 - 16i^{2}} = \frac{5 + 27i - 28}{1 + 16}$$
$$= \frac{-23 + 27i}{17} = -\frac{23}{17} + \frac{27}{17}i$$
Now find the argument of $j$:
$$\arg(j) = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{\frac{27}{17}}{-\frac{23}{17}}\right)$$
$$= \tan^{-1}\left(-\frac{27}{23}\right)$$
Since the real part is negative and the imaginary part is positive, $j$ is in the second quadrant. Therefore, we add $\pi$:
$$\arg(j) = \tan^{-1}\left(-\frac{27}{23}\right) + \pi$$
$$= \pi - \tan^{-1}\left(\frac{27}{23}\right)$$
$$\boxed{\arg(j) = \pi - \tan^{-1}\left(\frac{27}{23}\right) \text{ radians}}$$
(C) Find $h + j$:
$$h + j = \left(\frac{36}{13} - \frac{2}{13}i\right) + \left(-\frac{23}{17} + \frac{27}{17}i\right)$$
For the real parts (finding common denominator):
$$\frac{36}{13} - \frac{23}{17} = \frac{36 \cdot 17 - 23 \cdot 13}{13 \cdot 17} = \frac{612 - 299}{221} = \frac{313}{221}$$
For the imaginary parts:
$$-\frac{2}{13} + \frac{27}{17} = \frac{-2 \cdot 17 + 27 \cdot 13}{13 \cdot 17} = \frac{-34 + 351}{221} = \frac{317}{221}$$
Therefore:
$$\boxed{h + j = \frac{313}{221} + \frac{317}{221}i}$$
๐ Summary
๐ฏ Key Takeaways
- To find the conjugate of a complex number, flip the sign.
- To divide complex numbers, multiply numerator and denominator by the denominator's conjugate.
- To divide complex numbers by a real number, simply divide both the real and imaginary parts by the real number.
๐ What's Next?
Now that you understand how to divide complex numbers, you can explore more advanced topics. I suggest doing some past paper questions to strengthen your understanding, or doing modulus and argument of complex numbers.
Click here to learn the modulus and argument of complex numbers: