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Unity just means 1, so we need to find all z such that z^{3}=1. However, although we can immediately see that z=1 is a solution, when dealing with complex numbers some solution may not lie on the real axis. To find all the soltuions, it is necessary to re-write the equation in polar form, that is z^{3}=e^{2i*pi}, since 1 has argument 0 on the complex plane, and argument 2pi is the same as argument 0, as both will lie on the positive real axis. Now, we recall that if we add 2pi to the argument, the value is unchanged, since spinning a point 2pi around the origin in the complex plane puts the point back where it started, as 2pi is a full turn. Also, recall that cubic equations have 3 roots, so we can write z^{3}=e^{2i*pi}, e^{4i*pi} or e^{6i*pi}. Now, we can take cube roots on both sides, and recall that this is equivalent to raising to the power third, and that when raising a power to a power we multiply the powers. So we obtain z=e^{2i/3*pi}, e^{4I/3*pi} or e^{6i/3*pi}. Recall that we can subtract 2pi from the argument to ensure that the argument is between minus pi and pi, and that 6/3*pi=2pi which is the same as an argument of 0, i.e. this value is one. So the cube roots of unity are 1, e^{-1i/3*pi} and e^{2i/3*pi}.

Answered by Ethan T.

Studies Mathematics at Birmingham

Using the auxiliary equation t^2 - 2t - 3t = 0 t therefore is equal to 3 or -1. Using this value, a complementary function is derived. Y= Ae^(3x) + Be^(-x). Finally, to fully solve, a particular integral of y = asinx + bcosx and differentiate it twice, to give equations for Dy/dx and (d2y/dx2). These can be substituted into the initial differential equations to find the values of a and b, Which are -2/5 and 1/5 respectively.
The answer is then the complementary function plus the solution to the particular integral y = Ae^(3x) + Be^(-x) + (1/5)cosx - (2/5)sinx

Draw argand diagram
Modulus=|z|= length of the line
So |z|=(x^{2}+y^{2})^{0.5}
Argument=Angle between real (x axis) and line
so arg(z)=arctan(y/x)

Answered by Emma F.

Studies Aeronautics and Astronautics at Southampton

Before this proof, it is important to appreciate that both of these hyperbolic functions can be written in terms of e^x. Therefore, before you begin to differentiate, you must represent coshx as (e^x + e^-x)/2.
Then, this can easily be differentiated to give you the answer (e^x-e^-x)/2, which is the equivalent of sinhx.

Answered by Thomasina R.

Studies Physics at Exeter

What we should know is that the root 3+4i is a complex number that looks alot like a+bi.
We can say : rt(3+4i) = a+bi (Where we dont know what a & b is..yet)
and when we square both sides (rt(3+4i))^2=(a+bi)^2 | 3+4i = (a+bi)^2
we get 3+4i = a^2+2abi-b^2
We seperate the Real and Imaginary parts to get a simultainus equation
3 = a^2-b^2
4 = 2ba
if this is solved we get a= (+-)2 and b =(+-)1
to get (+-)(2+i) <--- which is the answer

This is an example of an integral that uses trigonometric substitutions, which is quite a common theme for A-level further maths questions. Firstly, notice the denominator is quite unpleasant, so to make things easier we make a substitution. The best substitution here would be to let x = sin(u). Why is this the case? Well, notice that this way the denominator becomes sqrt(1 - sin^{2}(u)) = sqrt(cos^{2}(u)) = cos(u) which is much tidier. We can now calculate dx with respect to this new substitution as well. Since x = sin(u), dx/du = cos(u) and so dx = cos(u)du. Thus after substitution, our integral becomes cos(u)/cos(u)du = the integral of 1 du = u + k where k is the constant of integration. Substituting back into terms of x, notice u = sin^{-1}(x), and so the entire integral is equal to sin^{-1}(x) + k. In general, when dealing with integration questions that have something of a nasty form, especially including square roots, there's probably a substitution that makes everything work out nicely. Be sure you have as many trigonometric identities memorised as possible, and try to select the best one to manipulate the integral into something workable.

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