Sapewa

2021-12-28

Having the following complex numbers $A=5-2i$ and $B=3+8i$
a) $A+B$
b) $B-A$
c) $A×B$
d) What is the complex conjugate of A and the complex conjugate of B?
e) $\frac{B}{A}$ (B divide by A)

Durst37

Step 1
a) Given complex numbers
$A=5-2i$
$B=3+8i$
$\left(a+ib\right)+\left(c+id\right)=\left(a+c\right)+i\left(b+d\right)$
Step 2
so the sum of given numbers
$A+B=\left(5-2i\right)+\left(3+8i\right)$
$=\left(5+3\right)+i\left(-2+8\right)$
$=8+6i$
Step 3
b) In subtraction of complex numbers real terms are subtracted and imaginary terms are subtracted together.
$\left(a+ib\right)-\left(c+id\right)=\left(a-c\right)+i\left(b-d\right)$
Step 4
so the difference of the terms
$B-A=\left(3+8i\right)-\left(5-2i\right)$
$=\left(3-5\right)+i\left(8+2\right)$
$=-2+10i$
Step 5
c)The product of two complex numbers can be found by
$\left(a+ib\right)×\left(c+id\right)=\left(ac-bd\right)+i\left(ad+bc\right)$
Step 6
Here the product is
$A×B=\left(5-2i\right)×\left(3+8i\right)$
$=\left(5×3-\left(-2\right)×8\right)+i\left(8×5+\left(-2\right)×3\right)$
$=31+34i$
Step 7
d)The complex conjugate of
$a+ib$ is $a-ib$
$a-ib$ is $a+ib$
Step 8
The complex conjugate of
$A=5-2i$
$⇒\stackrel{―}{A}=5+2i$
$B=3+8i$
$⇒\stackrel{―}{B}=3-8i$
Step 9
e) In complex numbers for division multiply the numerator and denominator by conjugate of denominator
$\frac{a+ib}{c+id}=\frac{a+ib}{c+id}×\frac{c-id}{c-id}$
$=\frac{\left(ac-bd\right)+i\left(ad+bc\right)}{{c}^{2}+{d}^{2}}$
Step 10
Here the division of B by A will give
$\frac{B}{A}=\frac{\left(3+8i\right)\left(5+2i\right)}{\left(5-2i\right)\left(5+2i\right)}$

peterpan7117i

Step 1
a) $\left(5-2i\right)+\left(3+8i\right)$
$=\left(5+3\right)+\left(-2+8\right)i$
$5+3=8$
$-2+8=6$
$=8+6i$
Step 2
b) $\left(3+8i\right)-\left(5-2i\right)$
$=\left(3-5\right)+\left(8+2\right)i$
$3-5=-2$
$8+2=10$
$=-2+10i$
Step 3
c) $\left(5-2i\right)×\left(3+8i\right)$
$=\left(5×3-\left(-2\right)×8\right)+\left(5×8+\left(-2\right)×3\right)i$
$5×3-\left(-2\right)×8=31$
$5×8+\left(-2\right)×3=34$
$=31+34i$
Step 4
e) $\frac{5-2i}{3+8i}$
Apply complex arithmetric rule:
$\frac{a+bi}{c+di}=\frac{\left(c-di\right)\left(a+bi\right)}{\left(c-di\right)\left(c+di\right)}=\frac{\left(ac+bd\right)+\left(bc-ad\right)i}{{c}^{2}+{a}^{2}}$

$=\frac{\left(5×3+\left(-2\right)×8\right)+\left(-2×3-5×8\right)i}{{3}^{2}+{8}^{2}}$
Refine
$=\frac{-1-46i}{73}$
Rewrite $\frac{-1-46i}{73}$ in standard complex form: $-\frac{1}{73}-\frac{46}{73}i$
$=-\frac{1}{73}-\frac{46}{73}i$

Vasquez

Step 1
a) Your input: simplify and calculate different forms of $3+5-2i+8i$
Rewrite: $3+5-2i+8i=8+6i$
Step 2
b) $3-\left(5-2i\right)+8i$
Rewrite: $3-\left(5-2i\right)+8i=-2+10i$
Step 3
c) $\left(3+8i\right)×\left(5-2i\right)$
Use FOIL to multiply (for steps, see foil calculator), don't forget that ${i}^{2}=-1$
$\left(\left(3+8i\right)×\left(5-2i\right)\right)=\left(31+34i\right)$
Hence, $\left(3+8i\right)×\left(5-2i\right)=31+34i$
Step 4
e) $\frac{5-2i}{3+8i}$
Rewrite: $\frac{5-2i}{3+8i}=\frac{\left(3-8i\right)\left(5-2i\right)}{73}$
Multiply using the formula $a\left(b+c\right)=ab+ac$
$\frac{3-8i}{73}=\left(\frac{1}{73}\right)\left(3\right)+\left(\frac{1}{73}\right)\left(-8i\right)=\frac{3}{73}-\frac{8i}{73}$
So,
$\left(\frac{3-8i}{73}\right)\left(5-2i\right)=\left(\frac{3}{73}-\frac{8i}{73}\right)\left(5-2i\right)$
Use FOIL to multiply (for steps, see foil calculator), don't forget that ${i}^{2}=-1$
$\left(\left(\frac{3}{73}-\frac{8i}{73}\right)\left(5-2i\right)\right)=\left(-\frac{1}{73}-\frac{46i}{73}\right)$
Hence, $\frac{\left(3-8i\right)\left(5-2i\right)}{73}=-\frac{1}{73}-\frac{46i}{73}$

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