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Here we use the complex exponential form of 1 which is e^(i 2n pi). Applying the sixth root and substituting in for integer values of n gives all roots in complex exponential form.These can be converted i...

The simplest way to describe a complex number is by its real and imaginary part, z=x+yi, this may be wrote as Re(z)=x and Im(z)=y. These complex numbers follow the same r...

First we remember that sinθ can be expressed in terms of powers of z, where z=cos(θ)+isin(θ), using the following:2isin(nθ)=zⁿ-z⁻ⁿ and 2cos(nθ)=zⁿ+z⁻ⁿ
so, [2isin(θ)]⁴=[z¹-z⁻¹]⁴ 16sin(θ)=(z)⁴(-z⁻¹)⁰+4...

First check that this works for n=1:2^(2x1 - 1) + 3^(2x1 - 1) = 2^1 +3^1 = 5 (so true for n=1)Now we assume this to work for any n = k.Assumption: 2^(2k-1) + 3^(2k-1) = 5a, where a is some integer constan...

Consider the general equationdy/dx + Py = Qwhere P and Q are functions of x.R (which will be introduced later) is also a function of x.
So, all of a sudden, we are going to just state the product rul...