Solve frac{5+sqrt{2}i}{1-sqrt{2}i}

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

Evaluate

$$1+2\sqrt{2}i\approx 1+2.828427125i$$

Solution Steps

Rationalize the denominator of \frac{5+i\sqrt{2}}{1-i\sqrt{2}} by multiplying numerator and denominator by 1+i\sqrt{2}.

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

Consider \left(1-i\sqrt{2}\right)\left(1+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}.

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

Calculate 1 to the power of 2 and get 1.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -1 and -2 to get 2.

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

Add 1 and 2 to get 3.

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

Apply the distributive property by multiplying each term of 5+i\sqrt{2} by each term of 1+i\sqrt{2}.

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

Combine 5i\sqrt{2} and i\sqrt{2} to get 6i\sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Subtract 2 from 5 to get 3.

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

Real Part

$$1$$