Our equation looks the same as the original, except C is now different. Put it together and we get Ax^2 + Bx + (C/i) = 0. Now, let's try dividing the whole equation by "i": even though it's imaginary, it's still a number, right? iAx^2 becomes Ax^2, iBx becomes Bx, C becomes (C/i), and 0 becomes 0 (0 divided by ANY number, even imaginary ones, is always equal to 0). Think about it! So if you have some quadratic equation Ax^2 + Bx + C = 0, and you multiply each "x" term by i, you'd get iAx^2 + iBx C = 0. Note: x+2pi*k just refers to any angle coterminal with x, so they are equivalent so what about the +2pi*k? Well, substitute x with x+2pi*k and voila! (In that first case, n represents any old number, but with 1/n, n is the number of roots) Keeping that in mind, let's arbitrarily substitute n for 1/n. Technically (cos x + i*sin x)^n = (r*(cos x + i*sin x))^n, but the r is omitted because we're usually talking about the unit circle (r=1). Clearly a very different formula, but that doesn't mean it can't be derived from that first expression. I don't know if this is how you've seen it, but the version of de Moivre's theorem in my textbook is r^(1/n)*(cos((x + 2pi*k)/n) + i*sin((x + 2pi*k)/n)). I can't seem to find any videos on de Moivre's theorem, either, but I do know that the idea that (cos x + i*sin x)^n = cos(nx) + i*sin(nx) can be derived from Euler's formula. The red line is the parabola that you normally get when using only real coordinates Plot3(realPart, zeros(1,numberOfPointsOnTheAxis). ImaginaryPart(:) * ones(1,numberOfPointsOnTheAxis) * 1i X = ones(numberOfPointsOnTheAxis,1) * realPart(:)' +. ImaginaryPart = linspace(-2,2,numberOfPointsOnTheAxis) RealPart = linspace(1,4,numberOfPointsOnTheAxis) P = % 1*x*x -6*x +10 the coefficients of the polynomial I came up with the following piece of code written in MATLAB: You could make two representations, one for the real value of the result and one for the imaginary value of the result, but you would have to search for the point(s) where those 2 are both 0. The absolute value is always non-negative, and the solutions to the polynomial are located at the points where the absolute value of the result is 0. You would put the absolute value of the result on the z-axis when x is real (complex part is 0) the absolute value is equal to the value of the polynomial at that point. So we finish our factors with a minus sign in front of the ?1? and a plus sign in front of the ?2?.I think a way to do that is to make a 3D chart that has the complex coordinates on the horizontal axis. But remember that in ?3x^2+5x-2?, the last term ?-2? is negative, which means one of our signs has to be negative, so the only two possibilities areīut neither of these is correct because we don’t get the ?+5x? in the middle. We need to combine ?3x? and ?2x? in such a way that we get ?5x?. Let’s see what happens if we use the first way. The only factors of ?2? are ?2? and ?1?, which means we’ll have one of the following. The only factors of ?3? are ?3? and ?1?, so we know we'll have Let’s begin by looking at the factors of ?3? and ?2?. Examples of factoring quadratics with coefficients
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