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Subtracting Complex Numbers – Techniques, Explanation, and Examples
![Subtracting Complex Numbers](https://www.storyofmathematics.com/wp-content/uploads/2023/03/Subtracting-Complex-Numbers-Title.png)
How to subtract complex numbers?
The general form of a complex number is of the form $a + bi$. If we have another complex number, $m + ni$, we can find the difference between the two by following the steps below:- Distribute the negative sign inside the parenthesis.
- Group the real number and the imaginary number parts.
- Simplify each group by adding or subtracting the coefficients.
![How to subtract complex numbers?](https://www.storyofmathematics.com/wp-content/uploads/2023/03/Subtracting-Complex-Numbers.png)
Examples of complex number subtraction
The degree of difficulty when subtracting complex numbers may vary, so we’ll show some common cases you may encounter.Case 1: Subtracting a complex number from a real numberLet’s say we have the real number, $m$. We can subtract $a + bi$ from it by distributing the negative sign before the parenthesis correctly.$ \begin{aligned} m – (a + bi) &= m – 1(a) – 1(bi)\\&= m – a – bi\end{aligned}$We can apply a similar process whenever we want to subtract a complex number from the given real number.Case 2: Subtracting a complex number from an imaginary numberLet’s say we have an imaginary number, $ni$. We can subtract $a + bi$ from it by:- Distributing the negative sign correctly into the parenthesis.
- Combine the two imaginary numbers.
Example 1
What is the result when $-8$ is subtracted from $-5 + 6i$? SolutionWe want to find the difference between $-5 + 6i$ and $-8$. Make sure to distribute the negative sign correctly. Combine the real number of parts to simplify the result.$ \begin{aligned} -5 + 6i – (-8) &= -5 + 6i + 8\\&= (-5 + 8) + 6i\\ &= 3 + 6i\end{aligned}$This means that when $-8$ is subtracted from $-5 + 6i$, the result is $3 + 6i$.![Subtracting complex numbers Example](https://www.storyofmathematics.com/wp-content/uploads/2023/03/Subtracting-Complex-Numbers-Example.png)
Example 2
What is the result when $12 – 8i$ is subtracted from $4i$? SolutionWe’re looking for the value of $4i – (12 – 8i)$. Distributing the negative sign properly is a crucial part of this problem.Combine the real and imaginary number parts, then simplify the result further, as shown below.$ \begin{aligned} 4i – (12 – 8i) &= 4 – 1(12) – (-8i)\\&= 4 + 12 + 8i\\&= (4 + 12) + 8i\\&=16 + 8i\end{aligned}$Hence, when $12 – 8i$ is subtracted from $4i$, the result is $16+ 8i$.Example 3
What is the result when $15- 3i$ is subtracted from $6i + 12$? SolutionMake sure that we take note which of the two complex numbers are placed before and after the word “from,” so we’re actually looking for $(6i + 12) – (15 – 13i)$. Rearrange the terms in the first group so that the real number goes first.Distribute the negative sign after then combine like terms.$ \begin{aligned} (6i + 12) – (15 – 13i) &= (12 + 6i) – ( 15 – 13i)\\&= 12 + 6i – 15 – 13i\end{aligned}$We can then group the real and imaginary number parts then simplify the resulting difference.$ \begin{aligned} 12 + 6i – 15 – 13i &= (12 – 15) + (6 – 13)i\\&= 3 – 3i\end{aligned}$This shows that when $15- 3i$ is subtracted from $6i + 12$, the result is $3 – 3i$.Example 4
Evaluate and simplify the following expressions.a. $(4 – 2i) – (3 + 5i) – (-6 – 9i)$ b. $[(12i – 8) – (10 – 4i)] – [(-6i – 15) – (8 + 10i)]$SolutionLet’s start with the first expression and simultaneously distribute the negative signs into the second and third sgroups of complex numbers.$\begin{aligned} (4 – 2i) – (3 + 5i) – (-6 – 9i) &= 4- 2i -(3) – (5i) -(-6) – (-9i)\\&= 4- 2i – 3 – 5i + 6 + 9i \end{aligned}$Now, let’s group all the real numbers and the imaginary numbers. Simplify each group to find the final value of the expression.$\begin{aligned}4- 2i – 3 – 5i + 6 + 9i &= (4 – 3 +6) + (-2i – 5i + 9i)\\&= (4 – 3 + 6) + (-2 – 5 + 9)i\\&= 7 + 2i \end{aligned}$a. This means that $(4 – 2i) – (3 + 5i) – (-6 – 9i)$ is equal to $7 + 2i$.We’re working with four complex numbers grouped into two subgroups. What we can do is find the difference between the two complex numbers in each bracket.Some reminders to remember:- Make sure that the complex number is of the form $a + bi$.
- Distribute the negative signs carefully to avoid any mistake at the first stage.
- Simplify by combining like the real and imaginary numbers.
$\boldsymbol{[(12i – 8) – (10 – 4i)]}$ | $\boldsymbol{[(-6i – 15) – (8 + 10i)]}$ |
$\begin{aligned}(12i – 8) – (10 – 4i) &= (-8 + 12i) – (10 – 4i)\\&= -8 + 12i – 10 + 4i\\&= (-8 – 10) + (12 + 4)i\\&= -18 + 16i\end{aligned}$ | $\begin{aligned}(-6i – 15) – (8 + 10i) &= (-15 – 6i) – (8 + 10i)\\&= -15 – 6i – 8 – 10i\\&= (-15 – 8) + (-6 -10)i\\&= -23 -16i \end{aligned}$ |
Example 5
Evaluate and simplify $[(4 + \sqrt{-36}) – (-3 – \sqrt{-49})] – [(-6 – \sqrt{-16}) – (12 – \sqrt{-81})]$.SolutionWe can see four square roots that contain negative values inside, so let’s rewrite these terms first in terms of $i$.- $\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i$
- $\sqrt{-49} = \sqrt{49} \cdot \sqrt{-1} = 7i$
- $\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i$
- $\sqrt{-81} = \sqrt{81} \cdot \sqrt{-1} = 9i$
Example 6
Show that the commutative property does not apply to the subtraction of complex numbers. SolutionLet $a + bi$ and $m + ni$ be the two complex numbers we want to subtract. To prove that subtraction of two complex numbers is not commutative, we want to show that $(a + bi) – (m + ni) \neq (m + ni) – (a + bi)$.We can evaluate the difference between the two expressions shown on the left and the right-hand side. We can apply the following steps to make sure we get the right expressions:- Distribute the negative signs correctly.
- Group the real and imaginary number parts.
- Simplify the expressions to find the difference of the two.
$\boldsymbol{(a + bi) – (m + ni)}$ | $\boldsymbol{(m + ni) – (a + bi)}$ |
$\begin{aligned} (a + bi) – (m + ni) &= a + bi -m – ni\\&= (a – m) + (bi – ni)\\&= (a – m) + (b -n)i\end{aligned}$ | $\begin{aligned} (m + ni) – (a + bi) &= m + ni – a- bi\\&= (m – a) + (ni – bi)\\&= (m – a) + (n -b)i\end{aligned}$ |
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Practice Questions
Open Problem
Show that the associative property does not apply to the subtraction of complex numbers.
Open Problem Solution
$\begin{aligned}(a+ bi)-[(c+di) -(e +fi)]&= (a -c +e) +(b – d +f)i\\(c+di)-[(a+bi)-(e+fi)]&=(c-a+e)+(d-b+f)i\\(a+ bi)-[(c+di) -(e +fi)]&\neq(c+di)-[(a+bi)-(e+fi)]\end{aligned}$