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

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

Evaluate

$$\frac{7}{2}-\frac{7}{2}i=3.5-3.5i$$

Solution Steps

Consider \left(3+i\sqrt{5}\right)\left(3-i\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{9-\left(i\sqrt{5}\right)^{2}}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)}$$

Expand \left(i\sqrt{5}\right)^{2}.

$$\frac{9-i^{2}\left(\sqrt{5}\right)^{2}}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)}$$

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

$$\frac{9-\left(-\left(\sqrt{5}\right)^{2}\right)}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)}$$

The square of \sqrt{5} is 5.

$$\frac{9-\left(-5\right)}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)}$$

The opposite of -5 is 5.

$$\frac{9+5}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)}$$

Add 9 and 5 to get 14.

$$\frac{14}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)}$$

Calculate the square root of 2i and get 1+i.

$$\frac{14}{\sqrt{3}+\left(1+i\right)-\left(\sqrt{3}-\sqrt{2i}\right)}$$

Calculate the square root of 2i and get 1+i.

$$\frac{14}{\sqrt{3}+\left(1+i\right)-\left(\sqrt{3}-\left(1+i\right)\right)}$$

Multiply -1 and 1+i to get -1-i.

$$\frac{14}{\sqrt{3}+\left(1+i\right)-\left(\sqrt{3}+\left(-1-i\right)\right)}$$

To find the opposite of \sqrt{3}+\left(-1-i\right), find the opposite of each term.

$$\frac{14}{\sqrt{3}+\left(1+i\right)-\sqrt{3}+\left(1+i\right)}$$

Combine the real and imaginary parts in 1+i+\left(1+i\right).

$$\frac{14}{\sqrt{3}-\sqrt{3}+1+1+\left(1+1\right)i}$$

Do the additions in 1+1+\left(1+1\right)i.

$$\frac{14}{\sqrt{3}-\sqrt{3}+2+2i}$$

Combine \sqrt{3} and -\sqrt{3} to get 0.

$$\frac{14}{2+2i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-2i.

$$\frac{14\left(2-2i\right)}{\left(2+2i\right)\left(2-2i\right)}$$

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

$$\frac{14\left(2-2i\right)}{2^{2}-2^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{14\left(2-2i\right)}{8}$$

Multiply 14 times 2-2i.

$$\frac{14\times 2+14\times \left(-2i\right)}{8}$$

Do the multiplications in 14\times 2+14\times \left(-2i\right).

$$\frac{28-28i}{8}$$

Divide 28-28i by 8 to get \frac{7}{2}-\frac{7}{2}i.

$$\frac{7}{2}-\frac{7}{2}i$$

Real Part

$$\frac{7}{2} = 3\frac{1}{2} = 3.5$$

Solution Steps

Consider \left(3+i\sqrt{5}\right)\left(3-i\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$Re(\frac{9-\left(i\sqrt{5}\right)^{2}}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)})$$

Expand \left(i\sqrt{5}\right)^{2}.

$$Re(\frac{9-i^{2}\left(\sqrt{5}\right)^{2}}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)})$$

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

$$Re(\frac{9-\left(-\left(\sqrt{5}\right)^{2}\right)}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)})$$

The square of \sqrt{5} is 5.

$$Re(\frac{9-\left(-5\right)}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)})$$

The opposite of -5 is 5.

$$Re(\frac{9+5}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)})$$

Add 9 and 5 to get 14.

$$Re(\frac{14}{\sqrt{3}+\sqrt{2i}-\left(\sqrt{3}-\sqrt{2i}\right)})$$

Calculate the square root of 2i and get 1+i.

$$Re(\frac{14}{\sqrt{3}+\left(1+i\right)-\left(\sqrt{3}-\sqrt{2i}\right)})$$

Calculate the square root of 2i and get 1+i.

$$Re(\frac{14}{\sqrt{3}+\left(1+i\right)-\left(\sqrt{3}-\left(1+i\right)\right)})$$

Multiply -1 and 1+i to get -1-i.

$$Re(\frac{14}{\sqrt{3}+\left(1+i\right)-\left(\sqrt{3}+\left(-1-i\right)\right)})$$

To find the opposite of \sqrt{3}+\left(-1-i\right), find the opposite of each term.

$$Re(\frac{14}{\sqrt{3}+\left(1+i\right)-\sqrt{3}+\left(1+i\right)})$$

Combine the real and imaginary parts in 1+i+\left(1+i\right).

$$Re(\frac{14}{\sqrt{3}-\sqrt{3}+1+1+\left(1+1\right)i})$$

Do the additions in 1+1+\left(1+1\right)i.

$$Re(\frac{14}{\sqrt{3}-\sqrt{3}+2+2i})$$

Combine \sqrt{3} and -\sqrt{3} to get 0.

$$Re(\frac{14}{2+2i})$$

Multiply both numerator and denominator of \frac{14}{2+2i} by the complex conjugate of the denominator, 2-2i.

$$Re(\frac{14\left(2-2i\right)}{\left(2+2i\right)\left(2-2i\right)})$$

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

$$Re(\frac{14\left(2-2i\right)}{2^{2}-2^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{14\left(2-2i\right)}{8})$$

Multiply 14 times 2-2i.

$$Re(\frac{14\times 2+14\times \left(-2i\right)}{8})$$

Do the multiplications in 14\times 2+14\times \left(-2i\right).

$$Re(\frac{28-28i}{8})$$

Divide 28-28i by 8 to get \frac{7}{2}-\frac{7}{2}i.

$$Re(\frac{7}{2}-\frac{7}{2}i)$$

The real part of \frac{7}{2}-\frac{7}{2}i is \frac{7}{2}.

$$\frac{7}{2}$$