Solve 2i/7+i

$$\frac { 2 i } { 7 + i }$$

Evaluate

$$\frac{1}{25}+\frac{7}{25}i=0.04+0.28i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 7-i.

$$\frac{2i\left(7-i\right)}{\left(7+i\right)\left(7-i\right)}$$

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

$$\frac{2i\left(7-i\right)}{7^{2}-i^{2}}$$

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

$$\frac{2i\left(7-i\right)}{50}$$

Multiply 2i times 7-i.

$$\frac{2i\times 7+2\left(-1\right)i^{2}}{50}$$

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

$$\frac{2i\times 7+2\left(-1\right)\left(-1\right)}{50}$$

Do the multiplications in 2i\times 7+2\left(-1\right)\left(-1\right). Reorder the terms.

$$\frac{2+14i}{50}$$

Divide 2+14i by 50 to get \frac{1}{25}+\frac{7}{25}i.

$$\frac{1}{25}+\frac{7}{25}i$$

Real Part

$$\frac{1}{25} = 0.04$$

Solution Steps

Multiply both numerator and denominator of \frac{2i}{7+i} by the complex conjugate of the denominator, 7-i.

$$Re(\frac{2i\left(7-i\right)}{\left(7+i\right)\left(7-i\right)})$$

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

$$Re(\frac{2i\left(7-i\right)}{7^{2}-i^{2}})$$

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

$$Re(\frac{2i\left(7-i\right)}{50})$$

Multiply 2i times 7-i.

$$Re(\frac{2i\times 7+2\left(-1\right)i^{2}}{50})$$

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

$$Re(\frac{2i\times 7+2\left(-1\right)\left(-1\right)}{50})$$

Do the multiplications in 2i\times 7+2\left(-1\right)\left(-1\right). Reorder the terms.

$$Re(\frac{2+14i}{50})$$

Divide 2+14i by 50 to get \frac{1}{25}+\frac{7}{25}i.

$$Re(\frac{1}{25}+\frac{7}{25}i)$$

The real part of \frac{1}{25}+\frac{7}{25}i is \frac{1}{25}.

$$\frac{1}{25}$$