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

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

Evaluate

$$\frac{3}{2}+\frac{1}{2}i=1.5+0.5i$$

Solution Steps

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

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

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

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

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

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

Combine the real and imaginary parts in 1-2i+2i+4.

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

Do the additions in 1+4+\left(-2+2\right)i.

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

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

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

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

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

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

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

Multiply 5 times 3+i.

$$\frac{5\times 3+5i}{10}$$

Do the multiplications in 5\times 3+5i.

$$\frac{15+5i}{10}$$

Divide 15+5i by 10 to get \frac{3}{2}+\frac{1}{2}i.

$$\frac{3}{2}+\frac{1}{2}i$$

Real Part

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

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

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

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

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

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

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

Combine the real and imaginary parts in 1-2i+2i+4.

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

Do the additions in 1+4+\left(-2+2\right)i.

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

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

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

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

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

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

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

Multiply 5 times 3+i.

$$Re(\frac{5\times 3+5i}{10})$$

Do the multiplications in 5\times 3+5i.

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

Divide 15+5i by 10 to get \frac{3}{2}+\frac{1}{2}i.

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

The real part of \frac{3}{2}+\frac{1}{2}i is \frac{3}{2}.

$$\frac{3}{2}$$