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

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

Evaluate

$$\frac{3}{10}-\frac{11}{10}i=0.3-1.1i$$

Solution Steps

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

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

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

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

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

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

Combine the real and imaginary parts in 3-3i+2i+2.

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

Do the additions in 3+2+\left(-3+2\right)i.

$$\frac{5-i}{2+4i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-4i.

$$\frac{\left(5-i\right)\left(2-4i\right)}{\left(2+4i\right)\left(2-4i\right)}$$

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

$$\frac{\left(5-i\right)\left(2-4i\right)}{2^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(5-i\right)\left(2-4i\right)}{20}$$

Multiply complex numbers 5-i and 2-4i like you multiply binomials.

$$\frac{5\times 2+5\times \left(-4i\right)-i\times 2-\left(-4i^{2}\right)}{20}$$

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

$$\frac{5\times 2+5\times \left(-4i\right)-i\times 2-\left(-4\left(-1\right)\right)}{20}$$

Do the multiplications in 5\times 2+5\times \left(-4i\right)-i\times 2-\left(-4\left(-1\right)\right).

$$\frac{10-20i-2i-4}{20}$$

Combine the real and imaginary parts in 10-20i-2i-4.

$$\frac{10-4+\left(-20-2\right)i}{20}$$

Do the additions in 10-4+\left(-20-2\right)i.

$$\frac{6-22i}{20}$$

Divide 6-22i by 20 to get \frac{3}{10}-\frac{11}{10}i.

$$\frac{3}{10}-\frac{11}{10}i$$

Real Part

$$\frac{3}{10} = 0.3$$

Solution Steps

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

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

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

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

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

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

Combine the real and imaginary parts in 3-3i+2i+2.

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

Do the additions in 3+2+\left(-3+2\right)i.

$$Re(\frac{5-i}{2+4i})$$

Multiply both numerator and denominator of \frac{5-i}{2+4i} by the complex conjugate of the denominator, 2-4i.

$$Re(\frac{\left(5-i\right)\left(2-4i\right)}{\left(2+4i\right)\left(2-4i\right)})$$

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

$$Re(\frac{\left(5-i\right)\left(2-4i\right)}{2^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(5-i\right)\left(2-4i\right)}{20})$$

Multiply complex numbers 5-i and 2-4i like you multiply binomials.

$$Re(\frac{5\times 2+5\times \left(-4i\right)-i\times 2-\left(-4i^{2}\right)}{20})$$

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

$$Re(\frac{5\times 2+5\times \left(-4i\right)-i\times 2-\left(-4\left(-1\right)\right)}{20})$$

Do the multiplications in 5\times 2+5\times \left(-4i\right)-i\times 2-\left(-4\left(-1\right)\right).

$$Re(\frac{10-20i-2i-4}{20})$$

Combine the real and imaginary parts in 10-20i-2i-4.

$$Re(\frac{10-4+\left(-20-2\right)i}{20})$$

Do the additions in 10-4+\left(-20-2\right)i.

$$Re(\frac{6-22i}{20})$$

Divide 6-22i by 20 to get \frac{3}{10}-\frac{11}{10}i.

$$Re(\frac{3}{10}-\frac{11}{10}i)$$

The real part of \frac{3}{10}-\frac{11}{10}i is \frac{3}{10}.

$$\frac{3}{10}$$