Solve 2+i+left(1+iright)left(4-3iright)/3+2i

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

Evaluate

$$\frac{31}{13}-\frac{12}{13}i\approx 2.384615385-0.923076923i$$

Solution Steps

Multiply complex numbers 1+i and 4-3i like you multiply binomials.

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

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

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

Do the multiplications in 1\times 4+1\times \left(-3i\right)+4i-3\left(-1\right).

$$\frac{2+i+4-3i+4i+3}{3+2i}$$

Combine the real and imaginary parts in 4-3i+4i+3.

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

Do the additions in 4+3+\left(-3+4\right)i.

$$\frac{2+i+\left(7+i\right)}{3+2i}$$

Combine the real and imaginary parts in 2+i+7+i.

$$\frac{2+7+\left(1+1\right)i}{3+2i}$$

Do the additions in 2+7+\left(1+1\right)i.

$$\frac{9+2i}{3+2i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 3-2i.

$$\frac{\left(9+2i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}$$

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

$$\frac{\left(9+2i\right)\left(3-2i\right)}{3^{2}-2^{2}i^{2}}$$

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

$$\frac{\left(9+2i\right)\left(3-2i\right)}{13}$$

Multiply complex numbers 9+2i and 3-2i like you multiply binomials.

$$\frac{9\times 3+9\times \left(-2i\right)+2i\times 3+2\left(-2\right)i^{2}}{13}$$

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

$$\frac{9\times 3+9\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right)}{13}$$

Do the multiplications in 9\times 3+9\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right).

$$\frac{27-18i+6i+4}{13}$$

Combine the real and imaginary parts in 27-18i+6i+4.

$$\frac{27+4+\left(-18+6\right)i}{13}$$

Do the additions in 27+4+\left(-18+6\right)i.

$$\frac{31-12i}{13}$$

Divide 31-12i by 13 to get \frac{31}{13}-\frac{12}{13}i.

$$\frac{31}{13}-\frac{12}{13}i$$

Real Part

$$\frac{31}{13} = 2\frac{5}{13} = 2.3846153846153846$$

Solution Steps

Multiply complex numbers 1+i and 4-3i like you multiply binomials.

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

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

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

Do the multiplications in 1\times 4+1\times \left(-3i\right)+4i-3\left(-1\right).

$$Re(\frac{2+i+4-3i+4i+3}{3+2i})$$

Combine the real and imaginary parts in 4-3i+4i+3.

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

Do the additions in 4+3+\left(-3+4\right)i.

$$Re(\frac{2+i+\left(7+i\right)}{3+2i})$$

Combine the real and imaginary parts in 2+i+7+i.

$$Re(\frac{2+7+\left(1+1\right)i}{3+2i})$$

Do the additions in 2+7+\left(1+1\right)i.

$$Re(\frac{9+2i}{3+2i})$$

Multiply both numerator and denominator of \frac{9+2i}{3+2i} by the complex conjugate of the denominator, 3-2i.

$$Re(\frac{\left(9+2i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)})$$

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

$$Re(\frac{\left(9+2i\right)\left(3-2i\right)}{3^{2}-2^{2}i^{2}})$$

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

$$Re(\frac{\left(9+2i\right)\left(3-2i\right)}{13})$$

Multiply complex numbers 9+2i and 3-2i like you multiply binomials.

$$Re(\frac{9\times 3+9\times \left(-2i\right)+2i\times 3+2\left(-2\right)i^{2}}{13})$$

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

$$Re(\frac{9\times 3+9\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right)}{13})$$

Do the multiplications in 9\times 3+9\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right).

$$Re(\frac{27-18i+6i+4}{13})$$

Combine the real and imaginary parts in 27-18i+6i+4.

$$Re(\frac{27+4+\left(-18+6\right)i}{13})$$

Do the additions in 27+4+\left(-18+6\right)i.

$$Re(\frac{31-12i}{13})$$

Divide 31-12i by 13 to get \frac{31}{13}-\frac{12}{13}i.

$$Re(\frac{31}{13}-\frac{12}{13}i)$$

The real part of \frac{31}{13}-\frac{12}{13}i is \frac{31}{13}.

$$\frac{31}{13}$$