Solve 1/-7+6i=

$$\frac { 1 } { - 7 + 6 i } =$$

Evaluate

$$-\frac{7}{85}-\frac{6}{85}i\approx -0.082352941-0.070588235i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -7-6i.

$$\frac{1\left(-7-6i\right)}{\left(-7+6i\right)\left(-7-6i\right)}$$

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

$$\frac{1\left(-7-6i\right)}{\left(-7\right)^{2}-6^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{1\left(-7-6i\right)}{85}$$

Multiply 1 and -7-6i to get -7-6i.

$$\frac{-7-6i}{85}$$

Divide -7-6i by 85 to get -\frac{7}{85}-\frac{6}{85}i.

$$-\frac{7}{85}-\frac{6}{85}i$$

Real Part

$$-\frac{7}{85} = -0.08235294117647059$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{-7+6i} by the complex conjugate of the denominator, -7-6i.

$$Re(\frac{1\left(-7-6i\right)}{\left(-7+6i\right)\left(-7-6i\right)})$$

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

$$Re(\frac{1\left(-7-6i\right)}{\left(-7\right)^{2}-6^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{1\left(-7-6i\right)}{85})$$

Multiply 1 and -7-6i to get -7-6i.

$$Re(\frac{-7-6i}{85})$$

Divide -7-6i by 85 to get -\frac{7}{85}-\frac{6}{85}i.

$$Re(-\frac{7}{85}-\frac{6}{85}i)$$

The real part of -\frac{7}{85}-\frac{6}{85}i is -\frac{7}{85}.

$$-\frac{7}{85}$$