Solve (1+i)^-1+(-2/3)^-1+3-i/1-sqrt{2i}+sqrt{2}i^19

$$( 1 + i ) ^ { - 1 } + ( - \frac { 2 } { 3 } ) ^ { - 1 } + \frac { 3 - i } { 1 - \sqrt { 2 } i } + \sqrt { 2 } i ^ { 19 }$$

Evaluate

$$\frac{\sqrt{2}}{3}-\frac{5}{6}i\approx 0.471404521-0.833333333i$$

Solution Steps

Calculate 1+i to the power of -1 and get \frac{1}{2}-\frac{1}{2}i.

$$\frac{1}{2}-\frac{1}{2}i+\left(-\frac{2}{3}\right)^{-1}+\frac{3-i}{1-i\sqrt{2}}+\sqrt{2}i^{19}$$

Calculate -\frac{2}{3} to the power of -1 and get -\frac{3}{2}.

$$\frac{1}{2}-\frac{1}{2}i-\frac{3}{2}+\frac{3-i}{1-i\sqrt{2}}+\sqrt{2}i^{19}$$

Do the additions.

$$\frac{3-i}{1-i\sqrt{2}}+\sqrt{2}i^{19}-1-\frac{1}{2}i$$

Calculate i to the power of 19 and get -i.

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

Rationalize the denominator of \frac{3-i}{1-i\sqrt{2}} by multiplying numerator and denominator by 1+i\sqrt{2}.

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

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}.

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

Calculate 1 to the power of 2 and get 1.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -1 and -2 to get 2.

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

Add 1 and 2 to get 3.

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

Divide \left(3-i\right)\left(1+i\sqrt{2}\right) by 3 to get \left(1-\frac{1}{3}i\right)\left(1+i\sqrt{2}\right).

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

Use the distributive property to multiply 1-\frac{1}{3}i by 1+i\sqrt{2}.

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

Do the additions.

$$\left(\frac{1}{3}+i\right)\sqrt{2}+\sqrt{2}\left(-i\right)-\frac{5}{6}i$$

Combine \left(\frac{1}{3}+i\right)\sqrt{2} and \sqrt{2}\left(-i\right) to get \frac{1}{3}\sqrt{2}.

$$\frac{1}{3}\sqrt{2}-\frac{5}{6}i$$

Real Part

$$\frac{\sqrt{2}}{3} = 0.47140452079103173$$