Review complex number addition, subtraction, and multiplication.
Addition
(a1+b1i)+(a2+b2i)=(a1+a2)+(b1+b2)i(a_1+b_1i)+(a_2+b_2i)=(a_1+a_2)+(b_1+b_2)i
Subtraction
(a1+b1i)(a2+b2i)=(a1a2)+(b1b2)i(a_1+b_1i)-(a_2+b_2i)=(a_1-a_2)+(b_1-b_2)i
Multiplication
(a1+b1i)(a2+b2i)=(a1a2b1b2)+(a1b2+a2b1)i(a_1+b_1i)\cdot(a_2+b_2i)=(a_1a_2-b_1b_2)+(a_1b_2+a_2b_1)i
Want to learn more about complex number operations? Check out these videos:

Practice set 1: Adding and subtracting complex numbers

Example 1: Adding complex numbers

When adding complex numbers, we simply add the real parts and add the imaginary parts. For example:
=(3+4i)+(610i)=(3+6)+(410)i=96i\begin{aligned} &\phantom{=}(\blueD 3+\greenD4i)+(\blueD6\greenD{-10}i) \\\\ &=(\blueD3+\blueD6)+(\greenD4\greenD{-10})i \\\\ &=\blueD9\greenD{-6}i \end{aligned}

Example 2: Subtracting complex numbers

When subtracting complex numbers, we simply subtract the real parts and subtract the imaginary parts. For example:
=(3+4i)(610i)=(36)+(4(10))i=3+14i\begin{aligned} &\phantom{=}(\blueD 3+\greenD4i)-(\blueD6\greenD{-10}i) \\\\ &=(\blueD3-\blueD6)+(\greenD4-(\greenD{-10}))i \\\\ &=\blueD{-3}+\greenD{14}i \end{aligned}
Want to try more problems like this? Check out this exercise.

Practice set 2: Multiplying complex numbers

When multiplying complex numbers, we perform a multiplication similar to how we expand the parentheses in binomial products:
(a+b)(c+d)=ac+ad+bc+bd(a+b)(c+d)=ac+ad+bc+bd
Unlike regular binomial multiplication, with complex numbers we also consider the fact that i2=1i^2=-1.

Example 1

=2(3+4i)=2(3)+24i=6+8i\begin{aligned} &\phantom{=}\blueD 2\cdot(\blueD{-3}+\greenD{4}i) \\\\ &=\blueD2\cdot(\blueD{-3})+\blueD2\cdot\greenD4i \\\\ &=\blueD{-6}+\greenD8i \end{aligned}

Example 2

=3i(15i)=3i1+3i(5)i=3i15i2=3i15(1)=15+3i\begin{aligned} &\phantom{=}\greenD3i\cdot(\blueD{1}\greenD{-5}i) \\\\ &=\greenD3i\cdot\blueD1+\greenD3i\cdot(\greenD{-5})i \\\\ &=\greenD3i-15i^2 \\\\ &=\greenD3i-15(-1) \\\\ &=\blueD{15}+\greenD3i \end{aligned}

Example 3

=(2+3i)(15i)=21+2(5)i+3i1+3i(5)i=210i+3i15i2=27i15(1)=177i\begin{aligned} &\phantom{=}(\blueD2+\greenD3i)\cdot(\blueD{1}\greenD{-5}i) \\\\ &=\blueD2\cdot\blueD1+\blueD2\cdot(\greenD{-5})i+\greenD3i\cdot\blueD1+\greenD3i\cdot(\greenD{-5})i \\\\ &=\blueD2\greenD{-10}i+\greenD3i-15i^2 \\\\ &=\blueD2\greenD{-7}i-15(-1) \\\\ &=\blueD{17}\greenD{-7}i \end{aligned}
Want to try more problems like this? Check out this basic exercise and this advanced exercise.
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