Solve 6+i/-6-2i

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

Evaluate

$$-\frac{19}{20}+\frac{3}{20}i=-0.95+0.15i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -6+2i.

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

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

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

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

$$\frac{\left(6+i\right)\left(-6+2i\right)}{40}$$

Multiply complex numbers 6+i and -6+2i like you multiply binomials.

$$\frac{6\left(-6\right)+6\times \left(2i\right)-6i+2i^{2}}{40}$$

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

$$\frac{6\left(-6\right)+6\times \left(2i\right)-6i+2\left(-1\right)}{40}$$

Do the multiplications in 6\left(-6\right)+6\times \left(2i\right)-6i+2\left(-1\right).

$$\frac{-36+12i-6i-2}{40}$$

Combine the real and imaginary parts in -36+12i-6i-2.

$$\frac{-36-2+\left(12-6\right)i}{40}$$

Do the additions in -36-2+\left(12-6\right)i.

$$\frac{-38+6i}{40}$$

Divide -38+6i by 40 to get -\frac{19}{20}+\frac{3}{20}i.

$$-\frac{19}{20}+\frac{3}{20}i$$

Real Part

$$-\frac{19}{20} = -0.95$$

Solution Steps

Multiply both numerator and denominator of \frac{6+i}{-6-2i} by the complex conjugate of the denominator, -6+2i.

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

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

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

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

$$Re(\frac{\left(6+i\right)\left(-6+2i\right)}{40})$$

Multiply complex numbers 6+i and -6+2i like you multiply binomials.

$$Re(\frac{6\left(-6\right)+6\times \left(2i\right)-6i+2i^{2}}{40})$$

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

$$Re(\frac{6\left(-6\right)+6\times \left(2i\right)-6i+2\left(-1\right)}{40})$$

Do the multiplications in 6\left(-6\right)+6\times \left(2i\right)-6i+2\left(-1\right).

$$Re(\frac{-36+12i-6i-2}{40})$$

Combine the real and imaginary parts in -36+12i-6i-2.

$$Re(\frac{-36-2+\left(12-6\right)i}{40})$$

Do the additions in -36-2+\left(12-6\right)i.

$$Re(\frac{-38+6i}{40})$$

Divide -38+6i by 40 to get -\frac{19}{20}+\frac{3}{20}i.

$$Re(-\frac{19}{20}+\frac{3}{20}i)$$

The real part of -\frac{19}{20}+\frac{3}{20}i is -\frac{19}{20}.

$$-\frac{19}{20}$$