Solve 5i/1-2i

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

Evaluate

$$-2+i$$

Solution Steps

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

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

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

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

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

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

Multiply 5i times 1+2i.

$$\frac{5i\times 1+5\times 2i^{2}}{5}$$

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

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

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

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

Divide -10+5i by 5 to get -2+i.

$$-2+i$$

Real Part

$$-2$$

Solution Steps

Multiply both numerator and denominator of \frac{5i}{1-2i} by the complex conjugate of the denominator, 1+2i.

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

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

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

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

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

Multiply 5i times 1+2i.

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

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

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

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

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

Divide -10+5i by 5 to get -2+i.

$$Re(-2+i)$$

The real part of -2+i is -2.

$$-2$$