kloseyq

2021-12-26

Find the sum and the product of the complex numbers $3-2i$ and $-3+5i$.

temzej9

Step 1
Consider the complex numbers $3-2i$ and $-3+5i$.
To find the sum of the complex numbers, add/subtract the real part with real and the imaginary part with the imaginary part.
So the sum of the complex numbers is
$\left(3-2i\right)+\left(-3+5i\right)=3-2i-3+5i$
$=3-3-2i+5i$
$=0+3i$
$=3i$
Thus, the sum of the complex numbers is 3i.
Step 2
To find the product of the complex number, use FOIL method to multiply the binomials and simplify further as shown below
$\left(3-2i\right)\left(-3+5i\right)=3×\left(-3\right)+3×\left(5i\right)-2i×\left(-3\right)-2i×\left(5i\right)$
$=-9+15i+6i-10{i}^{2}$
$=-9+21i-10\left(-1\right)\phantom{\rule{1em}{0ex}}\left({i}^{2}=-1\right)$
$=-9+21i+10$
$=1+21i$
Thus, the product of the complex numbers is $1+21i$

limacarp4

Step 1
Complex number:
1) $3-2i$
2) $-3+5i$
Add: $\left(1\right)+\left(2\right)=\left(3-2i\right)+\left(-3+5i\right)=\left(3-3\right)+\left(-2+5\right)i=3i$
Step 2
$\left(3-2i\right)\left(-3+5i\right)$
Apply complex arithmetric rule:
$\left(a+bi\right)\left(c+di\right)=\left(ac-bd\right)+\left(ad+bc\right)i$

$=\left(3\left(-3\right)-\left(-2\right)×5\right)+\left(3×5+\left(-2\right)\left(-3\right)\right)i$
$3\left(-3\right)-\left(-2\right)×5=1$
$3×5+\left(-2\right)\left(-3\right)=21$
$=1+21i$

Vasquez

$\left(3-2i\right)+\left(-3+5t\right)$
Group the real part and the imaginary part of the complex number

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