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

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

Evaluate

$$\frac{21}{5}+\frac{18}{5}i=4.2+3.6i$$

Solution Steps

Multiply both numerator and denominator of \frac{1+i}{2-i} by the complex conjugate of the denominator, 2+i.

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

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

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

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

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

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

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

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

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

Do the multiplications in 1\times 2+i+2i-1.

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

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

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

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

$$\frac{1+3i}{5}+\left(2-i\right)\left(1+2i\right)$$

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

$$\frac{1}{5}+\frac{3}{5}i+\left(2-i\right)\left(1+2i\right)$$

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

$$\frac{1}{5}+\frac{3}{5}i+2\times 1+2\times \left(2i\right)-i-2i^{2}$$

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

$$\frac{1}{5}+\frac{3}{5}i+2\times 1+2\times \left(2i\right)-i-2\left(-1\right)$$

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

$$\frac{1}{5}+\frac{3}{5}i+2+4i-i+2$$

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

$$\frac{1}{5}+\frac{3}{5}i+2+2+\left(4-1\right)i$$

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

$$\frac{1}{5}+\frac{3}{5}i+\left(4+3i\right)$$

Combine the real and imaginary parts.

$$\frac{1}{5}+4+\left(\frac{3}{5}+3\right)i$$

Do the additions.

$$\frac{21}{5}+\frac{18}{5}i$$

Real Part

$$\frac{21}{5} = 4\frac{1}{5} = 4.2$$

Solution Steps

Multiply both numerator and denominator of \frac{1+i}{2-i} by the complex conjugate of the denominator, 2+i.

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

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

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

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

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

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

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

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

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

Do the multiplications in 1\times 2+i+2i-1.

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

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

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

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

$$Re(\frac{1+3i}{5}+\left(2-i\right)\left(1+2i\right))$$

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

$$Re(\frac{1}{5}+\frac{3}{5}i+\left(2-i\right)\left(1+2i\right))$$

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

$$Re(\frac{1}{5}+\frac{3}{5}i+2\times 1+2\times \left(2i\right)-i-2i^{2})$$

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

$$Re(\frac{1}{5}+\frac{3}{5}i+2\times 1+2\times \left(2i\right)-i-2\left(-1\right))$$

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

$$Re(\frac{1}{5}+\frac{3}{5}i+2+4i-i+2)$$

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

$$Re(\frac{1}{5}+\frac{3}{5}i+2+2+\left(4-1\right)i)$$

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

$$Re(\frac{1}{5}+\frac{3}{5}i+\left(4+3i\right))$$

Combine the real and imaginary parts in \frac{1}{5}+\frac{3}{5}i+4+3i.

$$Re(\frac{1}{5}+4+\left(\frac{3}{5}+3\right)i)$$

Do the additions in \frac{1}{5}+4+\left(\frac{3}{5}+3\right)i.

$$Re(\frac{21}{5}+\frac{18}{5}i)$$

The real part of \frac{21}{5}+\frac{18}{5}i is \frac{21}{5}.

$$\frac{21}{5}$$