Not sure if its me or the book error but I had a complex number question
(6+i)(a+bi)=2
My solution for
My book has a different solution to that.
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I spent 5 minutes on this and couldn't solve it.
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
I think my book has done another mistake. (The book has made a few mistakes so I'm just checking with people on here)
| B Eng (Hons) | IB Mathematics SL | IB Mathematics HL | Australian Cricket | Casual University Statistics Tutor
i got an
x+yi
my x-component was :
I haven't solved the y-component yet. Want to get 'x' right before I do the y.
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The book's answer is right, you can check it by squaring it to get 5 + i.
x^2 - y^2 + 2xyi = 5+i
Comparing real and imaginary
2xy = 1
y = 1/2x
x^2 - y^2 = 5
x^2 - 1/4x^2 = 5
4x^4 - 20x^2 - 1 = 0
x^2 = (20 +- sqrt(20^2 - 4(4)(-1))/2(4)
x^2 = (20+- sqrt(416))/8
x^2 = (20 +- 4sqrt(26))/8
x^2 = (5 +- sqrt(26))/2
But x is real so take the positive case only
x = +- sqrt[(5+sqrt26)/2]
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