Solve 4+2i/1-2i

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

Evaluate

$$2i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$\frac{4+8i+2i-4}{5}$$

Combine the real and imaginary parts in 4+8i+2i-4.

$$\frac{4-4+\left(8+2\right)i}{5}$$

Do the additions in 4-4+\left(8+2\right)i.

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

Divide 10i by 5 to get 2i.

$$2i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{4+2i}{1-2i} by the complex conjugate of the denominator, 1+2i.

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

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

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

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

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

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

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

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

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

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

$$Re(\frac{4+8i+2i-4}{5})$$

Combine the real and imaginary parts in 4+8i+2i-4.

$$Re(\frac{4-4+\left(8+2\right)i}{5})$$

Do the additions in 4-4+\left(8+2\right)i.

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

Divide 10i by 5 to get 2i.

$$Re(2i)$$

The real part of 2i is 0.

$$0$$