![]() ![]() But since x^3 = i is degree three, there should be three different values of x. If you take i (sqrt(-1)), then the cube root is -i. ![]() If x^3 = N, where N is some expression (which could be a constant), then you have a degree three equation so there must be three roots. Complex cube roots, by DeMoivreĮnough of square roots let’s take it up a notch, with this question from 1997 on cube roots: Cube Roots of Numbers We could also do this using Euler or DeMoivre. We said above that a had to be real, so it must be the case that a = 2 or a = -2.Įach of the two solutions for x gives us two solutions for a, and each real value of a yields a corresponding value for b. This is quadratic in a^2 so we make a substitution: The substitution left us with a variable on the bottom, so we cleared that and got what is sometimes called a biquadratic equation. These are the same equations we saw before, but with different constants on the right, which make us do a little more work this time: Solve for b in the second equation and substitute into the first equation: Please help me.ĭoctor Paul answered, using the algebraic approach: If sqrt(3+4*i) = a + b*i where a and b are real numbers, then squaring both sides gives: I am in advanced college math and am totally stumped by this question. How about the square root of a complex number other than i? Here is a (poorly titled!) question from 2002: Finding the Square Root of a Quadratic Function But don’t forget to adjust \(\theta\) to be in the right quadrant. This amounts to the proof of DeMoivre we saw last time. Any number x + iy can be written in terms of a radius r and angle theta (counterclockwise from the x axis): Euler's formula makes it easy to find powers and roots by working in polar coordinates in the complex plane. We’ll start with three different perspectives on the same problem, \(\sqrt\) are the 90° and 450° we saw before. ![]() Now let’s look at roots of complex numbers, which will bring us to the idea that led to this series, namely the roots of unity. So, putting a and b back into the context of the question, we have two solutions: 5-2i and -5+2i.We’ve looked at positive integer powers of complex numbers, and imaginary powers of e. This means that when a=5, b=-2 and when a=-5, b=2. Substitute each a value into our earlier expression for b. This means our solutions are a=5 and a=-5. We have assumed a to be real so a 2=-4 has no solutions of interest to us. Some simplification and factorisation of this equation gives us (a 2+4)(a 2-25)=0, a quadratic in disguise. Now substitute this expression for b into equation 1. We can solve these simultaneous equations for a and b.įirstly, we can make b the subject of equation 2 by dividing both sides by 2a. We now have two equations in two unknowns. Next, let's compare the imaginary parts of the equation (the coefficients of i). We have a 2-b 2=21 (call this equation 1). Let's compare coeffiecients to obtain two equations in a and b.įirst, let's compare the real parts of the equation. Now both sides of the equation are in the same form. The natural step to take here is the mulitply out the term on the right-hand side.Īs i 2=-1 by definition of i, this equation can be rearranged to give 21-20i=(a 2-b 2)+(2ab)i. We also know that x can be expressed as a+bi (where a and b are real) since the square roots of a complex number are always complex. We know that all square roots of this number will satisfy the equation 21-20i=x 2 by definition of a square root. Let's consider the complex number 21-20i. However since we don't know how to deal with expressions such as √i we need to follow a specific method to find the square roots of a complex number. Every complex number has complex square roots. ![]()
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