Solve (1+i/1-i)^6+frac{sqrt{2}+sqrt{3}i}{sqrt{3}-sqrt{2}i}

$$( \frac { 1 + i } { 1 - i } ) ^ { 6 } + \frac { \sqrt { 2 } + \sqrt { 3 } i } { \sqrt { 3 } - \sqrt { 2 } i }$$

Evaluate

$$-1+i$$

Solution Steps

Multiply both numerator and denominator of \frac{1+i}{1-i} by the complex conjugate of the denominator, 1+i.

$$\left(\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{6}+\frac{\sqrt{2}+i\sqrt{3}}{\sqrt{3}-i\sqrt{2}}$$

Do the multiplications in \frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

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

Divide 2i by 2 to get i.

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

Calculate i to the power of 6 and get -1.

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

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}.

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

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}.

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

The square of \sqrt{3} is 3.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -1 and -2 to get 2.

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

Add 3 and 2 to get 5.

$$-1+\frac{\left(\sqrt{2}+i\sqrt{3}\right)\left(\sqrt{3}+i\sqrt{2}\right)}{5}$$

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

$$-1+\frac{i\left(\sqrt{2}\right)^{2}+i\left(\sqrt{3}\right)^{2}}{5}$$

The square of \sqrt{2} is 2.

$$-1+\frac{2i+i\left(\sqrt{3}\right)^{2}}{5}$$

The square of \sqrt{3} is 3.

$$-1+\frac{2i+3i}{5}$$

Add 2i and 3i to get 5i.

$$-1+\frac{5i}{5}$$

Divide 5i by 5 to get i.

$$-1+i$$

Real Part

$$-1$$

Solution Steps

Multiply both numerator and denominator of \frac{1+i}{1-i} by the complex conjugate of the denominator, 1+i.

$$Re(\left(\frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}\right)^{6}+\frac{\sqrt{2}+i\sqrt{3}}{\sqrt{3}-i\sqrt{2}})$$

Do the multiplications in \frac{\left(1+i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

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

Divide 2i by 2 to get i.

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

Calculate i to the power of 6 and get -1.

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

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}.

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

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}.

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

The square of \sqrt{3} is 3.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -1 and -2 to get 2.

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

Add 3 and 2 to get 5.

$$Re(-1+\frac{\left(\sqrt{2}+i\sqrt{3}\right)\left(\sqrt{3}+i\sqrt{2}\right)}{5})$$

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

$$Re(-1+\frac{i\left(\sqrt{2}\right)^{2}+i\left(\sqrt{3}\right)^{2}}{5})$$

The square of \sqrt{2} is 2.

$$Re(-1+\frac{2i+i\left(\sqrt{3}\right)^{2}}{5})$$

The square of \sqrt{3} is 3.

$$Re(-1+\frac{2i+3i}{5})$$

Add 2i and 3i to get 5i.

$$Re(-1+\frac{5i}{5})$$

Divide 5i by 5 to get i.

$$Re(-1+i)$$

The real part of -1+i is -1.

$$-1$$