Solve (5+i)(7-6i)/3+i

$$\frac { ( 5 + i ) ( 7 - 6 i ) } { 3 + i }$$

Evaluate

$$10-11i$$

Solution Steps

Multiply complex numbers 5+i and 7-6i like you multiply binomials.

$$\frac{5\times 7+5\times \left(-6i\right)+7i-6i^{2}}{3+i}$$

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

$$\frac{5\times 7+5\times \left(-6i\right)+7i-6\left(-1\right)}{3+i}$$

Do the multiplications in 5\times 7+5\times \left(-6i\right)+7i-6\left(-1\right).

$$\frac{35-30i+7i+6}{3+i}$$

Combine the real and imaginary parts in 35-30i+7i+6.

$$\frac{35+6+\left(-30+7\right)i}{3+i}$$

Do the additions in 35+6+\left(-30+7\right)i.

$$\frac{41-23i}{3+i}$$

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

$$\frac{\left(41-23i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)}$$

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

$$\frac{\left(41-23i\right)\left(3-i\right)}{3^{2}-i^{2}}$$

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

$$\frac{\left(41-23i\right)\left(3-i\right)}{10}$$

Multiply complex numbers 41-23i and 3-i like you multiply binomials.

$$\frac{41\times 3+41\left(-i\right)-23i\times 3-23\left(-1\right)i^{2}}{10}$$

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

$$\frac{41\times 3+41\left(-i\right)-23i\times 3-23\left(-1\right)\left(-1\right)}{10}$$

Do the multiplications in 41\times 3+41\left(-i\right)-23i\times 3-23\left(-1\right)\left(-1\right).

$$\frac{123-41i-69i-23}{10}$$

Combine the real and imaginary parts in 123-41i-69i-23.

$$\frac{123-23+\left(-41-69\right)i}{10}$$

Do the additions in 123-23+\left(-41-69\right)i.

$$\frac{100-110i}{10}$$

Divide 100-110i by 10 to get 10-11i.

$$10-11i$$

Real Part

$$10$$

Solution Steps

Multiply complex numbers 5+i and 7-6i like you multiply binomials.

$$Re(\frac{5\times 7+5\times \left(-6i\right)+7i-6i^{2}}{3+i})$$

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

$$Re(\frac{5\times 7+5\times \left(-6i\right)+7i-6\left(-1\right)}{3+i})$$

Do the multiplications in 5\times 7+5\times \left(-6i\right)+7i-6\left(-1\right).

$$Re(\frac{35-30i+7i+6}{3+i})$$

Combine the real and imaginary parts in 35-30i+7i+6.

$$Re(\frac{35+6+\left(-30+7\right)i}{3+i})$$

Do the additions in 35+6+\left(-30+7\right)i.

$$Re(\frac{41-23i}{3+i})$$

Multiply both numerator and denominator of \frac{41-23i}{3+i} by the complex conjugate of the denominator, 3-i.

$$Re(\frac{\left(41-23i\right)\left(3-i\right)}{\left(3+i\right)\left(3-i\right)})$$

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

$$Re(\frac{\left(41-23i\right)\left(3-i\right)}{3^{2}-i^{2}})$$

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

$$Re(\frac{\left(41-23i\right)\left(3-i\right)}{10})$$

Multiply complex numbers 41-23i and 3-i like you multiply binomials.

$$Re(\frac{41\times 3+41\left(-i\right)-23i\times 3-23\left(-1\right)i^{2}}{10})$$

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

$$Re(\frac{41\times 3+41\left(-i\right)-23i\times 3-23\left(-1\right)\left(-1\right)}{10})$$

Do the multiplications in 41\times 3+41\left(-i\right)-23i\times 3-23\left(-1\right)\left(-1\right).

$$Re(\frac{123-41i-69i-23}{10})$$

Combine the real and imaginary parts in 123-41i-69i-23.

$$Re(\frac{123-23+\left(-41-69\right)i}{10})$$

Do the additions in 123-23+\left(-41-69\right)i.

$$Re(\frac{100-110i}{10})$$

Divide 100-110i by 10 to get 10-11i.

$$Re(10-11i)$$

The real part of 10-11i is 10.

$$10$$