Solve ((4-3i)^-1+(-2+i)^-1)

$$( ( 4 - 3 i ) ^ { - 1 } + ( - 2 + i ) ^ { - 1 } )$$

Evaluate

$$-\frac{6}{25}-\frac{2}{25}i=-0.24-0.08i$$

Solution Steps

Calculate 4-3i to the power of -1 and get \frac{4}{25}+\frac{3}{25}i.

$$\frac{4}{25}+\frac{3}{25}i+\left(-2+i\right)^{-1}$$

Calculate -2+i to the power of -1 and get -\frac{2}{5}-\frac{1}{5}i.

$$\frac{4}{25}+\frac{3}{25}i+\left(-\frac{2}{5}-\frac{1}{5}i\right)$$

Add \frac{4}{25}+\frac{3}{25}i and -\frac{2}{5}-\frac{1}{5}i to get -\frac{6}{25}-\frac{2}{25}i.

$$-\frac{6}{25}-\frac{2}{25}i$$

Real Part

$$-\frac{6}{25} = -0.24$$

Solution Steps

Calculate 4-3i to the power of -1 and get \frac{4}{25}+\frac{3}{25}i.

$$Re(\frac{4}{25}+\frac{3}{25}i+\left(-2+i\right)^{-1})$$

Calculate -2+i to the power of -1 and get -\frac{2}{5}-\frac{1}{5}i.

$$Re(\frac{4}{25}+\frac{3}{25}i+\left(-\frac{2}{5}-\frac{1}{5}i\right))$$

Add \frac{4}{25}+\frac{3}{25}i and -\frac{2}{5}-\frac{1}{5}i to get -\frac{6}{25}-\frac{2}{25}i.

$$Re(-\frac{6}{25}-\frac{2}{25}i)$$

The real part of -\frac{6}{25}-\frac{2}{25}i is -\frac{6}{25}.

$$-\frac{6}{25}$$