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tekilochka [14]
3 years ago
10

Two complex numbers are represented by c + di and , where c, d, e, and f are positive real numbers. In what two quadrants may th

e product of these complex numbers lie? Explain your answer in complete sentences.
Mathematics
1 answer:
myrzilka [38]3 years ago
6 0

Answer: In quadrant 1 or quadrant 2.

Step-by-step explanation:

We have the numbers:

x = c + di

y = e + fi

where c, d, e and f are real positive numbers.

the product of these numbers is:

x*y = (c + di)*(e + fi) = c*e + c*fi + d*ei ´+d*f*i^2

x*y = c*e - d*f + (c*f + d*e)i

where I used that i^2 = -1

knowing that c,f, d and e are positive numbers, then the imaginary part of the product must be always positive.

For the real part, we have c*e - d*f, that can be positive o negative depending on the values of c, e, d, and f.

So we have that the product must lie always in one of the upper two quadrants, quadrant 1 or quadrant 2 because the imaginary part is always positive and the real part can be positive or negative.

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Step-by-step explanation:

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Create an equivalent system of these equations and test your solution <br> x + y = 1<br> x 3y =9
ad-work [718]
An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides


assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent



ok, solve initial

x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation


-x-y=-1
<u>x-3y=9 +</u>
0x-4y=8

-4y=8
divide both sides by -4
y=-2

sub back
x+y=1
x-2=1
add 2
x=3


x=3
y=-2
(3,-2)

if we test it in other one

2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep

2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep


solution is (3,-2)
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