Solve 2i/5-i

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

Evaluate

$$-\frac{1}{13}+\frac{5}{13}i\approx -0.076923077+0.384615385i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5+i.

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

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

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

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

$$\frac{2i\left(5+i\right)}{26}$$

Multiply 2i times 5+i.

$$\frac{2i\times 5+2i^{2}}{26}$$

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

$$\frac{2i\times 5+2\left(-1\right)}{26}$$

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

$$\frac{-2+10i}{26}$$

Divide -2+10i by 26 to get -\frac{1}{13}+\frac{5}{13}i.

$$-\frac{1}{13}+\frac{5}{13}i$$

Real Part

$$-\frac{1}{13} = -0.07692307692307693$$

Solution Steps

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

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

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

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

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

$$Re(\frac{2i\left(5+i\right)}{26})$$

Multiply 2i times 5+i.

$$Re(\frac{2i\times 5+2i^{2}}{26})$$

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

$$Re(\frac{2i\times 5+2\left(-1\right)}{26})$$

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

$$Re(\frac{-2+10i}{26})$$

Divide -2+10i by 26 to get -\frac{1}{13}+\frac{5}{13}i.

$$Re(-\frac{1}{13}+\frac{5}{13}i)$$

The real part of -\frac{1}{13}+\frac{5}{13}i is -\frac{1}{13}.

$$-\frac{1}{13}$$