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

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

Evaluate

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

Solution Steps

Subtract 2 from 1 to get -1.

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

Calculate -1 to the power of 2 and get 1.

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

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

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

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

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

Square \sqrt{3}. Square i.

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

Subtract -1 from 3 to get 4.

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

Real Part

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