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

Sep 10, 2017

${\left(5 - \sqrt{3} i\right)}^{4} = 184 - 440 \sqrt{3} i$

#### 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, ${\left(5 - \sqrt{3} i\right)}^{4} = {5}^{4} - 4 \cdot {5}^{3} \cdot \sqrt{3} i + 6 \cdot {5}^{2} \cdot {\left(\sqrt{3} i\right)}^{2} - 4 \cdot 5 \cdot {\left(\sqrt{3} i\right)}^{3} + {\left(\sqrt{3} i\right)}^{4}$.

Simplifying this you get: ${\left(5 - \sqrt{3} i\right)}^{4} = 625 - 500 \sqrt{3} i - 450 + 60 \sqrt{3} i + 9$

Ultimately leading down to: ${\left(5 - \sqrt{3} i\right)}^{4} = 184 - 440 \sqrt{3} i$.

On a side note, this question is probably easier to attempt using de Moivre's theorem, which states that ${z}^{n} = {r}^{n} c i s \left(n \theta\right)$.

Let $z = 5 - \sqrt{3} i$, which must first be converted to polar form.

Therefore, $| z | = \sqrt{{5}^{2} + {\sqrt{3}}^{2}} = 2 \sqrt{7}$.

To find $A r g \left(z\right)$, evaluate $\arctan \left(- \frac{\sqrt{3}}{5}\right)$.

Therefore $z = 2 \sqrt{7} \cdot c i s \left(\arctan \left(- \frac{\sqrt{3}}{5}\right)\right)$

So, ${z}^{4} = 784 \cdot c i s \left(4 \arctan \left(- \frac{\sqrt{3}}{5}\right)\right)$

Then, re-convert back into Cartesian form.

$x = 784 \cdot \cos \left(4 \arctan \left(- \frac{\sqrt{3}}{5}\right)\right)$
$y = 784 \cdot \sin \left(4 \arctan \left(- \frac{\sqrt{3}}{5}\right)\right)$
$z = x + y i$.

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.