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

1 Answer
Ryan
Sep 10, 2017

(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.

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