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

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

Evaluate

$$\frac{-\sqrt{3}i-1}{4}\approx -0.25-0.433012702i$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+i\right)^{2}.

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

The square of \sqrt{3} is 3.

$$\frac{1-i\sqrt{3}}{3+2i\sqrt{3}-1}$$

Subtract 1 from 3 to get 2.

$$\frac{1-i\sqrt{3}}{2+2i\sqrt{3}}$$

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

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

Consider \left(2+2i\sqrt{3}\right)\left(2-2i\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{4-\left(2i\sqrt{3}\right)^{2}}$$

Expand \left(2i\sqrt{3}\right)^{2}.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{4-\left(2i\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2i to the power of 2 and get -4.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{4-\left(-4\left(\sqrt{3}\right)^{2}\right)}$$

The square of \sqrt{3} is 3.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{4-\left(-4\times 3\right)}$$

Multiply -4 and 3 to get -12.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{4-\left(-12\right)}$$

Multiply -1 and -12 to get 12.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{4+12}$$

Add 4 and 12 to get 16.

$$\frac{\left(1-i\sqrt{3}\right)\left(2-2i\sqrt{3}\right)}{16}$$

Use the distributive property to multiply 1-i\sqrt{3} by 2-2i\sqrt{3}.

$$\frac{2\left(1-i\sqrt{3}\right)-2i\left(1-i\sqrt{3}\right)\sqrt{3}}{16}$$

Use the distributive property to multiply 2 by 1-i\sqrt{3}.

$$\frac{2-2i\sqrt{3}-2i\left(1-i\sqrt{3}\right)\sqrt{3}}{16}$$

Use the distributive property to multiply -2i by 1-i\sqrt{3}.

$$\frac{2-2i\sqrt{3}+\left(-2i-2\sqrt{3}\right)\sqrt{3}}{16}$$

Use the distributive property to multiply -2i-2\sqrt{3} by \sqrt{3}.

$$\frac{2-2i\sqrt{3}-2i\sqrt{3}-2\left(\sqrt{3}\right)^{2}}{16}$$

The square of \sqrt{3} is 3.

$$\frac{2-2i\sqrt{3}-2i\sqrt{3}-2\times 3}{16}$$

Multiply -2 and 3 to get -6.

$$\frac{2-2i\sqrt{3}-2i\sqrt{3}-6}{16}$$

Combine -2i\sqrt{3} and -2i\sqrt{3} to get -4i\sqrt{3}.

$$\frac{2-4i\sqrt{3}-6}{16}$$

Subtract 6 from 2 to get -4.

$$\frac{-4-4i\sqrt{3}}{16}$$

Real Part

$$-\frac{1}{4} = -0.25$$