Question 1081232: Solve w^2 = 15+8i
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let w = a+bi be any complex number
Square both sides
w = a+bi
w^2 = (a+bi)^2
w^2 = (a+bi)(a+bi)
w^2 = a(a+bi)+bi(a+bi)
w^2 = a^2 + abi + abi + b^2i^2
w^2 = a^2 + abi + abi + b^2(-1)
w^2 = a^2 + abi + abi - b^2
w^2 = a^2 - b^2 + abi + abi
w^2 = (a^2 - b^2) + (2ab)i
Since w^2 = 15+8i, we can say
(a^2 - b^2) + (2ab)i = 15 + 8i
The real parts are (a^2-b^2) and 15 for the left and right sides respectively. This must mean they are equal, so,
a^2 - b^2 = 15
Likewise, the imaginary parts are (2ab)i and 8i for the left and right sides respectively. So,
(2ab)i = 8i
2ab = 8
2ab/2 = 8/2
ab = 4
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The system of equations we need to solve is
a^2 - b^2 = 15
ab = 4
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Let's solve equation (2) shown above for the variable b
ab = 4
ab/a = 4/a
b = 4/a
Which is then plugged into the other equation and we solve for 'a'
a^2 - b^2 = 15
a^2 - (4/a)^2 = 15
a^2 - 16/a^2 = 15
a^2*(a^2 - 16/a^2) = a^2*15
a^4 - 16 = 15a^2
a^4 - 16-15a^2 = 15a^2-15a^2
a^4-15a^2- 16 = 0
(a^2-16)(a^2+1) = 0
(a-4)(a+4)(a^2+1) = 0
a-4 = 0 or a+4 = 0 or a^2+1 = 0
a = 4 or a = -4 or a = i or a = -i
If a = 4, then
b = 4/a
b = 4/4
b = 1
Which means,
w = a+bi
w = 4+1i
w = 4+i
is one solution
If a = -4, then
b = 4/a
b = 4/(-4)
b = -1
Which means,
w = a+bi
w = -4-1i
w = -4-i
is another solution
If a = i, then
b = 4/a
b = 4/i
b = (4/i)*(i/i)
b = (4i)/(i^2)
b = (4i)/(-1)
b = -4i
Which means
w = a+bi
w = i+(-4i)i
w = i-4i^2
w = i-4(-1)
w = i+4
w = 4+i
is another solution, but as shown earlier, this solution occurs when a = 4
If a = -i, then
b = 4/a
b = 4/i
b = (4/(-i))*(i/i)
b = (4i)/(-i^2)
b = (4i)/(-(-1))
b = 4i
Which means
w = a+bi
w = -i+(4i)i
w = -i+4i^2
w = -i+4(-1)
w = -i-4
w = -4-i
but this is another repeat of what was shown when a = -4
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To summarize, the only two solutions are
w = 4+i
w = -4-i
To condense things a bit, we can write
)
note how -(4+i) = -4-i
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