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Solving Operations with Complex Numbers?

Solving Operations with Complex Numbers? Topic: Solving Operations with Complex Numbers?
June 16, 2019 / By Annmarie
Question: I need help figuring out how to do a problem with an exponent :) This inlcudes i (imaginary units) So the problem is: (2+3i)^2. I'm not sure if I go about doing (2+3i)(2+3i) and FOILing it, because the textbook says the answer is -5+12. I know that i^2 = -1. Either way, though, there is no example for exponents in the book. Thank you for your time.
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Best Answers: Solving Operations with Complex Numbers?

Yarwood Yarwood | 8 days ago
You're on the right track. FOIL and then apply the fact that i² = -1 (2+3i()2+3i) = 4 + 12i + 9i² = 4 + 12i + 9(-1) = 4 + 12i - 9 = -5 + 12i :)
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Yarwood Originally Answered: Help with Complex Numbers?
1) You are missing your exponent. ex. if your problem was x³ - 27=0 You would isolate the base to a power then take the cube root of 27 (what times itself times itself = 27 so ? x*x*x=27) so x³ = 27 then ∛x³ = ∛27 =∛3³ so x=3. 2.Take the square root twice or the fourth root once since the exponential term is already isolated. Hint 3*3*3*3=81 3. 8x³ - 12x² + 2x - 3=0 Split your P(x) into two parts then factor each. 4x²(2x - 3) +1(2x -3)=0 Now you have a common factor to factor out. (2x - 3)(4x² +1)=0 Now solve each factor by setting them =0 and solving for x. 2x - 3 = 0⇢⇢2x = 3⇢⇢x = 3/2 4x² +1 = 0⇢⇢4x² = -1⇢⇢x² = -1/4 taking the square roots gives x = ± i/2
Yarwood Originally Answered: Help with Complex Numbers?
1. x^(-27) = 0 x = 0^(-1/27) x = 27th root of (1/0) IMPOSSIBLE 2. x^4 = 81 x^2 = +/- sqrt(81) x^2 = +/- 9 x^2 = 9 or x^2 = -9 x = +/- sqrt(9) or x = +/- sqrt(-9) x = +/- 3 or x = +/- 3i 3. 8x^3 - 12x^2 + 2x - 3 = 0 4x^2(2x - 3) + (2x - 3) = 0 (4x^2 + 1)(2x - 3) = 0 4x^2 + 1 = 0 or 2x - 3 = 0 4x^2 = -1 or 2x = 3 x^2 = -1/4 or x = 3/2 x = +/- sqrt(-1/4) or x = 1.5 x = +/- i/2 or x = 1.5
Yarwood Originally Answered: Help with Complex Numbers?
x^3-27=0 (x-3)(x^2+3x+9)=0 x=3 a=1 b=3 c=9 -3+/- sq rt 3^2-4(1)(9)/ 2 -3+/- sq rt -27/2 -3+/- 3i sq rt 3/2 x=3 x=-3+ 3i / 2 x=-3-3i/2 x^4=81 x^4=3^4 x=+/- 3 2x-3)(4x^2+1)=0 x=-3/2 x= 4x^2=-1 x^2=-1/4 x= +1/4i -1/4i -3/2 1/4i -1/4i

Shelomi Shelomi
textbook says -5 plus 12i , possibly which you get if you multiply out your brackets. 4 + 6i +6i +9i^2 Where 9i^2 =-9
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Shelomi Originally Answered: imaginary and complex numbers?
x - 5i + 2 = 4 - 3i + yi x - 2 = 2i + yi x - 2 = i(2 + y) x = 2 + i(2 + y) x = 2 (non-imaginary part) i = 0/(2 + y) i(2 + y) = 0 Dividing both sides by i 2 + y = 0 y = -2
Shelomi Originally Answered: imaginary and complex numbers?
If you factor it into its real/imaginary parts then you can solve 2 simultaneous equations to find x and y, (x + 2) - 5j = 4 + (y - 3)j This means, (x +2 ) = 4 <- real part (y - 3) = -5 <- imaginary part Then it is just a case of solving the 2 equations.

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