Solve (frac{-2+1sqrt{3}}{sqrt{5}-6i})

$$( \frac { - 2 + 1 \sqrt { 3 } } { \sqrt { 5 } - 6 i } )$$

Evaluate

$$\frac{\left(\sqrt{3}-2\right)\left(\sqrt{5}+6i\right)}{41}\approx -0.014613478-0.039212077i$$

Solution Steps

Rationalize the denominator of \frac{-2+1\sqrt{3}}{\sqrt{5}-6i} by multiplying numerator and denominator by \sqrt{5}+6i.

$$\frac{\left(-2+1\sqrt{3}\right)\left(\sqrt{5}+6i\right)}{\left(\sqrt{5}-6i\right)\left(\sqrt{5}+6i\right)}$$

Consider \left(\sqrt{5}-6i\right)\left(\sqrt{5}+6i\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(-2+1\sqrt{3}\right)\left(\sqrt{5}+6i\right)}{\left(\sqrt{5}\right)^{2}-\left(-6i\right)^{2}}$$

Square \sqrt{5}. Square -6i.

$$\frac{\left(-2+1\sqrt{3}\right)\left(\sqrt{5}+6i\right)}{5+36}$$

Subtract -36 from 5 to get 41.

$$\frac{\left(-2+1\sqrt{3}\right)\left(\sqrt{5}+6i\right)}{41}$$

Apply the distributive property by multiplying each term of -2+1\sqrt{3} by each term of \sqrt{5}+6i.

$$\frac{-2\sqrt{5}-12i+1\sqrt{3}\sqrt{5}+6i\sqrt{3}}{41}$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{-2\sqrt{5}-12i+1\sqrt{15}+6i\sqrt{3}}{41}$$

For any term t, t\times 1=t and 1t=t.

$$\frac{-2\sqrt{5}-12i+\sqrt{15}+6i\sqrt{3}}{41}$$

Real Part

$$\frac{\sqrt{5} {(\sqrt{3} - 2)}}{41} = -0.014613478263223484$$