Solve (3+4i)(1+2i)/1+i=

$$\frac { ( 3 + 4 i ) ( 1 + 2 i ) } { 1 + i } =$$

Evaluate

$$\frac{5}{2}+\frac{15}{2}i=2.5+7.5i$$

Solution Steps

Multiply complex numbers 3+4i and 1+2i like you multiply binomials.

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

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

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

Do the multiplications in 3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2\left(-1\right).

$$\frac{3+6i+4i-8}{1+i}$$

Combine the real and imaginary parts in 3+6i+4i-8.

$$\frac{3-8+\left(6+4\right)i}{1+i}$$

Do the additions in 3-8+\left(6+4\right)i.

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

Multiply both numerator and denominator by the complex conjugate of the denominator, 1-i.

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

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

$$\frac{\left(-5+10i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

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

$$\frac{\left(-5+10i\right)\left(1-i\right)}{2}$$

Multiply complex numbers -5+10i and 1-i like you multiply binomials.

$$\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)i^{2}}{2}$$

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

$$\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right)}{2}$$

Do the multiplications in -5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right).

$$\frac{-5+5i+10i+10}{2}$$

Combine the real and imaginary parts in -5+5i+10i+10.

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

Do the additions in -5+10+\left(5+10\right)i.

$$\frac{5+15i}{2}$$

Divide 5+15i by 2 to get \frac{5}{2}+\frac{15}{2}i.

$$\frac{5}{2}+\frac{15}{2}i$$

Real Part

$$\frac{5}{2} = 2\frac{1}{2} = 2.5$$

Solution Steps

Multiply complex numbers 3+4i and 1+2i like you multiply binomials.

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

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

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

Do the multiplications in 3\times 1+3\times \left(2i\right)+4i\times 1+4\times 2\left(-1\right).

$$Re(\frac{3+6i+4i-8}{1+i})$$

Combine the real and imaginary parts in 3+6i+4i-8.

$$Re(\frac{3-8+\left(6+4\right)i}{1+i})$$

Do the additions in 3-8+\left(6+4\right)i.

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

Multiply both numerator and denominator of \frac{-5+10i}{1+i} by the complex conjugate of the denominator, 1-i.

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

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

$$Re(\frac{\left(-5+10i\right)\left(1-i\right)}{1^{2}-i^{2}})$$

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

$$Re(\frac{\left(-5+10i\right)\left(1-i\right)}{2})$$

Multiply complex numbers -5+10i and 1-i like you multiply binomials.

$$Re(\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)i^{2}}{2})$$

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

$$Re(\frac{-5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right)}{2})$$

Do the multiplications in -5-5\left(-i\right)+10i\times 1+10\left(-1\right)\left(-1\right).

$$Re(\frac{-5+5i+10i+10}{2})$$

Combine the real and imaginary parts in -5+5i+10i+10.

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

Do the additions in -5+10+\left(5+10\right)i.

$$Re(\frac{5+15i}{2})$$

Divide 5+15i by 2 to get \frac{5}{2}+\frac{15}{2}i.

$$Re(\frac{5}{2}+\frac{15}{2}i)$$

The real part of \frac{5}{2}+\frac{15}{2}i is \frac{5}{2}.

$$\frac{5}{2}$$