Question

How do you use the binomial theorem to expand `(5-sqrt3i)^4`?

Answer 1

`(5-sqrt(3)i)^4 = 184 -440sqrt(3)i`

Explanation:

Since you asked how to use the binomial theorem for this question, here goes:

The binomial theorem says that an expression of two terms added or subtracted raised to a power `n`, equals `"^n C_0*x^n*y^0`, adding terms so on and so forth with the power of `y` increasing as the power of `x` decreases, and the term being chosen approaches `n`.

Therefore, `(5-sqrt(3)i)^4 = 5^4 - 4*5^3*sqrt(3)i + 6*5^2*(sqrt(3)i)^2 - 4*5*(sqrt(3)i)^3 + (sqrt(3)i)^4`.

Simplifying this you get: `(5-sqrt(3)i)^4 = 625 - 500sqrt(3)i - 450 + 60sqrt(3)i + 9`

Ultimately leading down to: `(5-sqrt(3)i)^4 = 184 - 440sqrt(3)i`.

On a side note, this question is probably easier to attempt using de Moivre's theorem, which states that `z^n = r^n cis(ntheta)`.

Let `z = 5-sqrt(3)i`, which must first be converted to polar form.

Therefore, `|z| = sqrt(5^2 + sqrt(3)^2) = 2sqrt(7)`.

To find `Arg(z)`, evaluate `arctan(-sqrt(3)/5)`.

Therefore `z = 2sqrt(7) * cis (arctan(-sqrt(3)/5))`

So, `z^4 = 784 * cis (4arctan(-sqrt(3)/5))`

Then, re-convert back into Cartesian form.

`x= 784* cos(4arctan(-sqrt(3)/5))`
`y= 784* sin(4arctan(-sqrt(3)/5))`
`z=x + yi`.

Obviously this is only handy if you have a calculator on hand to crunch the numbers. You get the same result as above, but it's up to you which method you use.