SOLUTION: Please help me solve this question Given (4+2i)m - (1-i)n = -9+9i. Find the complex numbers m and n if m is the conjugate of n. thank you.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Please help me solve this question Given (4+2i)m - (1-i)n = -9+9i. Find the complex numbers m and n if m is the conjugate of n. thank you.      Log On


   

Question 1170874: Please help me solve this question
Given (4+2i)m - (1-i)n = -9+9i. Find the complex numbers m and n if m is the conjugate of n.
thank you.
Found 3 solutions by math_tutor2020, MathTherapy, Alan3354:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Let m and n be complex numbers
Furthermore, let n be the complex conjugate of m
This means
m = a+bi
n = a-bi
for real numbers a,b

Let's find an equivalent expression for (4+2i)m
(4+2i)m = (4+2i)(a+bi)
(4+2i)m = 4(a+bi)+2i(a+bi)
(4+2i)m = 4a+4bi+2ai+2bi^2
(4+2i)m = 4a+4bi+2ai+2b(-1)
(4+2i)m = 4a+4bi+2ai-2b
(4+2i) = (4a-2b)+(4bi+2ai)
(4+2i)m = (4a-2b)+(4b+2a)i
I'm using parenthesis to help separate out the real and imaginary parts.

Similarly, let's find (1-i)n
(1-i)n = (1-i)(a-bi)
(1-i)n = 1(a-bi)-i(a-bi)
(1-i)n = a-bi-ai+bi^2
(1-i)n = a-bi-ai+b(-1)
(1-i)n = a-bi-ai-b
(1-i)n = (a-b)+(-ai-bi)
(1-i)n = (a-b)-(ai+bi)
(1-i)n = (a-b)-(a+b)i

This means,
(4+2i)m-(1-i)n = [(4a-2b)+(4b+2a)i] - [(a-b)-(a+b)i]
(4+2i)m-(1-i)n = (4a-2b)+(4b+2a)i - (a-b)+(a+b)i
(4+2i)m-(1-i)n = [(4a-2b) - (a-b)] + [(4b+2a)i+(a+b)i]
(4+2i)m-(1-i)n = [(4a-2b) - (a-b)] + [(4b+2a)+(a+b)]i
(4+2i)m-(1-i)n = (4a-2b - a+b) + (4b+2a+a+b)i
(4+2i)m-(1-i)n = (3a-b) + (3a+5b)i


Equate that result with -9+9i
(3a-b) + (3a+5b)i = -9+9i
The corresponding parts must match up and be equal
  • 3a-b = -9
  • (3a+5b)i = 9i which leads to 3a+5b = 9

The system of equations is
3a-b = -9
3a+5b = 9

Let's subtract the terms straight down
3a-3a = 0a
-b-5b = -6b
-9-9 = -18

We end up with -6b = -18 which solves to b = 3 after dividing both sides by -6.

We can then find 'a'
3a-b = -9
3a-3 = -9
3a = -9+3
3a = -6
a = -6/3
a = -2

In summary we found that
a = -2
b = 3

which directly leads to
m = a+bi = -2+3i
n = a-bi = -2-3i

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Checking the answers:

(4+2i)m = (4+2i)(-2+3i)
(4+2i)m = 4(-2+3i)+2i(-2+3i)
(4+2i)m = -8+12i-4i+6i^2
(4+2i)m = -8+12i-4i+6(-1)
(4+2i)m = -8+12i-4i-6
(4+2i)m = (-8-6)+(12-4)i
(4+2i)m = -14+8i

(1-i)n = (1-i)(-2-3i)
(1-i)n = 1(-2-3i)-i(-2-3i)
(1-i)n = -2-3i+2i+3i^2
(1-i)n = -2-3i+2i+3(-1)
(1-i)n = -2-3i+2i-3
(1-i)n = (-2-3)+(-3+2)i
(1-i)n = -5-i

(4+2i)m-(1-i)n = (-14+8i) - (-5-i)
(4+2i)m-(1-i)n = -14+8i + 5+i
(4+2i)m-(1-i)n = (-14+5)+(8i+i)
(4+2i)m-(1-i)n = (-14+5)+(8+1)i
(4+2i)m-(1-i)n = -9+9i

The answers have been confirmed.
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Answers:
m = -2+3i
n = -2-3i

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Please help me solve this question
Given (4+2i)m - (1-i)n = -9+9i. Find the complex numbers m and n if m is the conjugate of n.
thank you.
(4 + 2i)m - (1  -  i)n = - 9 + 9i

Let m be a + bi
Then n is: a - bi, since n is the conjugate of m


We then get: (4 + 2i)(a + bi) - (1 - i)(a - bi) = - 9 + 9i


3a - b = - 9 ---- Equating 1st terms ------ eq (i)
3a + 5b = 9 ------- Equating 2nd terms ------ eq (ii)
     6b = 18 ------ Subtracting eq (i) from eq (ii)
     matrix%281%2C5%2C+b%2C+%22=%22%2C+18%2F6%2C+%22=%22%2C+3%29
3a - 3 = - 9 ------- Substituting 3 for b in eq (i)
    3a = - 6
    matrix%281%2C5%2C+a%2C+%22=%22%2C+%28-+6%29%2F3%2C+%22=%22%2C+-+2%29

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve this question
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