Solve frac{sqrt{2}-sqrt{3}i}{sqrt{3}+sqrt{2}i}+frac{sqrt{3}+sqrt{2}i}{sqrt{2}-sqrt{3}i}

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

Evaluate

$$0$$

Solution Steps

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -1 and -2 to get 2.

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

Add 3 and 2 to get 5.

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

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

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

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

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

The square of \sqrt{3} is 3.

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

Multiply -i and 3 to get -3i.

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

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

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

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

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

The square of \sqrt{2} is 2.

$$\frac{\sqrt{6}-3i-i\times 2-\sqrt{3}\sqrt{2}}{5}+\frac{\sqrt{3}+i\sqrt{2}}{\sqrt{2}-i\sqrt{3}}$$

Multiply -i and 2 to get -2i.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

Subtract 2i from -3i to get -5i.

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

Combine \sqrt{6} and -\sqrt{6} to get 0.

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

Divide -5i by 5 to get -i.

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

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

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

Consider \left(\sqrt{2}-i\sqrt{3}\right)\left(\sqrt{2}+i\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}.

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

The square of \sqrt{2} is 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -1 and -3 to get 3.

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

Add 2 and 3 to get 5.

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

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

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{3} is 3.

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

The square of \sqrt{2} is 2.

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

Add 3i and 2i to get 5i.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$-i+\frac{\sqrt{6}+5i-\sqrt{6}}{5}$$

Combine \sqrt{6} and -\sqrt{6} to get 0.

$$-i+\frac{5i}{5}$$

Divide 5i by 5 to get i.

$$-i+i$$

Add -i and i to get 0.

$$0$$

Real Part

$$0$$

Solution Steps

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -1 and -2 to get 2.

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

Add 3 and 2 to get 5.

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

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

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

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

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

The square of \sqrt{3} is 3.

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

Multiply -i and 3 to get -3i.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -i and 2 to get -2i.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

Subtract 2i from -3i to get -5i.

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

Combine \sqrt{6} and -\sqrt{6} to get 0.

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

Divide -5i by 5 to get -i.

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

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

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

Consider \left(\sqrt{2}-i\sqrt{3}\right)\left(\sqrt{2}+i\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}.

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

The square of \sqrt{2} is 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -1 and -3 to get 3.

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

Add 2 and 3 to get 5.

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

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

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{3} is 3.

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

The square of \sqrt{2} is 2.

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

Add 3i and 2i to get 5i.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$Re(-i+\frac{\sqrt{6}+5i-\sqrt{6}}{5})$$

Combine \sqrt{6} and -\sqrt{6} to get 0.

$$Re(-i+\frac{5i}{5})$$

Divide 5i by 5 to get i.

$$Re(-i+i)$$

Add -i and i to get 0.

$$Re(0)$$

The real part of 0 is 0.

$$0$$