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. Together you can come up with a plan to get you the help you need.I think a way to do that is to make a 3D chart that has the complex coordinates on the horizontal axis. See your instructor as soon as you can to discuss your situation. You should get help right away or you will quickly be overwhelmed. …no - I don’t get it! This is a warning sign and you must not ignore it. Is there a place on campus where math tutors are available? Can your study skills be improved? Who can you ask for help? Your fellow classmates and instructor are good resources. It is important to make sure you have a strong foundation before you move on. In math every topic builds upon previous work. This must be addressed quickly because topics you do not master become potholes in your road to success. What did you do to become confident of your ability to do these things? Be specific. Reflect on the study skills you used so that you can continue to use them. Congratulations! You have achieved the objectives in this section. Figure 9.1.23Ĭhoose how would you respond to the statement “I can solve quadratic equations of the form a times the square of \(x\) minus \(h\) equals \(k\) using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.” After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
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