Title: Course Presentation

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Multiplication of Complex Numbers

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The multiplication of complex numbers \(z_{1}=a+bi\) and \(z_{2}=c+di\) is done according to the binomial multiplication rule

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\(z_{1}\cdot z_{2} = (a+bi)\cdot(c+di)=ac+adi+bci+bdi^2=(ac-bd)-(ad+bc)i\)

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Here \(i^2=-1\) and \(bdi^2=-bd\)

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Example

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Calculate \((1-2i)\cdot(2+i)\)

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Using the binomial product, we get

\((1-2i)(2+i) = 1\cdot{2}+1\cdot{i}-2i\cdot{2}-2i\cdot{i} = 2+i-4i+2 = 4-3i\)

Another option is to find the product immediately through the last equation of the formula. Then

\((1-2i)(2+i) = (1\cdot{2}+2\cdot{1})+(1\cdot{1}-2\cdot{2})i = 4-3i\)

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Practice

Multiplication Fill in the Blanks

Calculate.

Q: 1) \((2+3i)\cdot(3-i) =\) [...]

2) \((1-4i)\cdot(-3+i)=\)[...]
A: 9+7i; 1+13i
Feedback (0 to 100%): Solution: 1)\((2+3i)\cdot(3-i) = 2\cdot3+2\cdot(-i)+3i\cdot3+3i\cdot(-i) = 6-2i+9i+3 = 9+7i\); 2)\((1-4i)(-3+i) = 1\cdot(-3)+1\cdot i-4i\cdot(-3)-4i\cdot i = -3+i+12i+4 = 1+13i\)