Solve (6+2i)-(1+3i)/(-1+i)-2

$$\frac { ( 6 + 2 i ) - ( 1 + 3 i ) } { ( - 1 + i ) - 2 }$$

Evaluate

$$-\frac{8}{5}-\frac{1}{5}i=-1.6-0.2i$$

Solution Steps

Subtract 1+3i from 6+2i by subtracting corresponding real and imaginary parts.

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

Subtract 1 from 6. Subtract 3 from 2.

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

Subtract 2 from -1+i by subtracting corresponding real and imaginary parts.

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

Subtract 2 from -1 to get -3.

$$\frac{5-i}{-3+i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, -3-i.

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

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

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

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

$$\frac{\left(5-i\right)\left(-3-i\right)}{10}$$

Multiply complex numbers 5-i and -3-i like you multiply binomials.

$$\frac{5\left(-3\right)+5\left(-i\right)-i\left(-3\right)-\left(-i^{2}\right)}{10}$$

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

$$\frac{5\left(-3\right)+5\left(-i\right)-i\left(-3\right)-\left(-\left(-1\right)\right)}{10}$$

Do the multiplications in 5\left(-3\right)+5\left(-i\right)-i\left(-3\right)-\left(-\left(-1\right)\right).

$$\frac{-15-5i+3i-1}{10}$$

Combine the real and imaginary parts in -15-5i+3i-1.

$$\frac{-15-1+\left(-5+3\right)i}{10}$$

Do the additions in -15-1+\left(-5+3\right)i.

$$\frac{-16-2i}{10}$$

Divide -16-2i by 10 to get -\frac{8}{5}-\frac{1}{5}i.

$$-\frac{8}{5}-\frac{1}{5}i$$

Real Part

$$-\frac{8}{5} = -1\frac{3}{5} = -1.6$$

Solution Steps

Subtract 1+3i from 6+2i by subtracting corresponding real and imaginary parts.

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

Subtract 1 from 6. Subtract 3 from 2.

$$Re(\frac{5-i}{-1+i-2})$$

Subtract 2 from -1+i by subtracting corresponding real and imaginary parts.

$$Re(\frac{5-i}{-1-2+i})$$

Subtract 2 from -1 to get -3.

$$Re(\frac{5-i}{-3+i})$$

Multiply both numerator and denominator of \frac{5-i}{-3+i} by the complex conjugate of the denominator, -3-i.

$$Re(\frac{\left(5-i\right)\left(-3-i\right)}{\left(-3+i\right)\left(-3-i\right)})$$

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

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

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

$$Re(\frac{\left(5-i\right)\left(-3-i\right)}{10})$$

Multiply complex numbers 5-i and -3-i like you multiply binomials.

$$Re(\frac{5\left(-3\right)+5\left(-i\right)-i\left(-3\right)-\left(-i^{2}\right)}{10})$$

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

$$Re(\frac{5\left(-3\right)+5\left(-i\right)-i\left(-3\right)-\left(-\left(-1\right)\right)}{10})$$

Do the multiplications in 5\left(-3\right)+5\left(-i\right)-i\left(-3\right)-\left(-\left(-1\right)\right).

$$Re(\frac{-15-5i+3i-1}{10})$$

Combine the real and imaginary parts in -15-5i+3i-1.

$$Re(\frac{-15-1+\left(-5+3\right)i}{10})$$

Do the additions in -15-1+\left(-5+3\right)i.

$$Re(\frac{-16-2i}{10})$$

Divide -16-2i by 10 to get -\frac{8}{5}-\frac{1}{5}i.

$$Re(-\frac{8}{5}-\frac{1}{5}i)$$

The real part of -\frac{8}{5}-\frac{1}{5}i is -\frac{8}{5}.

$$-\frac{8}{5}$$