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

1 Answer
Sep 10, 2017

Answer:

#(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.