Stefan Hendricks

2021-12-30

Explain how to multiply complex numbers and give an example.
$3+2i$ and $2-3i$

yotaniwc

Step 1
If the two complex numbers are $a+bi$ and $c+di$, the multiplication of these complex numbers is $\left(a+bi\right)\left(c+di\right)$
Simplify $\left(a+bi\right)\left(c+di\right)$ using distributive property and reduce thr powers of i's using ${i}^{2}=-1$
Step 2
For examples consider two complex numbers as $3+2i$ and $2-3i$
The multiplication of these two complex numbers is,
$\left(3+2i\right)\left(2-3i\right)=3×2+3\left(-3i\right)+\left(2i\right)×2+\left(2i\right)\left(-3i\right)$
$=6-9i+4i-6{i}^{2}$
$=6-5i-6\left(-1\right)$
$=6-5i+6$
$=12-5i$
Therefore, the multiplication of complex numbers $3+2i$ and $2-3i$ is $12-5i$

Stuart Rountree

Step 1
Given complex numbers: $3+2i$ and $2-3i$
Equation: $\left(3+2i\right)\left(2-3i\right)$
Apply complex arithmetric rule: $\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i$
$a=3$
$b=2$
$c=2$
$d=-3$
$=\left(3×2-2\left(-3\right)\right)+\left(3\left(-3\right)+2×2\right)i$
$3×2-2\left(-3\right)=12$
$3\left(-3\right)+2×2=-5$
$=12-5i$

Vasquez

Step 1
1) Complex number: 3+2i
2) Complex number: 2-3i
Multiple: $\text{the result of step 1}×\text{the result of step 2}$
=(3+2i)$×$(2-3i)
3$×$2+3$×$(-3i)+2i$×$2+2i$×$(-3i)
=6-9i+4i-6i${}^{2}$
=6-9i+4i+6
=6+6+i(-9+4)
=12-5i

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