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

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

Evaluate

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

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-i\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 1 to get -2.

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

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

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

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}.

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

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

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -4 and 3 to get -12.

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

Multiply -1 and -12 to get 12.

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

Add 4 and 12 to get 16.

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

Use the distributive property to multiply \sqrt{3}+i by -2+2i\sqrt{3} and combine like terms.

$$\frac{-4\sqrt{3}+2i\left(\sqrt{3}\right)^{2}-2i}{16}$$

The square of \sqrt{3} is 3.

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

Multiply 2i and 3 to get 6i.

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

Subtract 2i from 6i to get 4i.

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

Real Part

$$-\frac{\sqrt{3}}{4} = -0.4330127018922193$$