Solve 1+2i/1-2i

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

Evaluate

$$-\frac{3}{5}+\frac{4}{5}i=-0.6+0.8i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+2i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide -3+4i by 5 to get -\frac{3}{5}+\frac{4}{5}i.

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

Real Part

$$-\frac{3}{5} = -0.6$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide -3+4i by 5 to get -\frac{3}{5}+\frac{4}{5}i.

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

The real part of -\frac{3}{5}+\frac{4}{5}i is -\frac{3}{5}.

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