Solve 5:(7-2i)

$$5 : ( 7 - 2 i )$$

Evaluate

$$\frac{35}{53}+\frac{10}{53}i\approx 0.660377358+0.188679245i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 7+2i.

$$\frac{5\left(7+2i\right)}{\left(7-2i\right)\left(7+2i\right)}$$

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

$$\frac{5\left(7+2i\right)}{7^{2}-2^{2}i^{2}}$$

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

$$\frac{5\left(7+2i\right)}{53}$$

Multiply 5 times 7+2i.

$$\frac{5\times 7+5\times \left(2i\right)}{53}$$

Do the multiplications in 5\times 7+5\times \left(2i\right).

$$\frac{35+10i}{53}$$

Divide 35+10i by 53 to get \frac{35}{53}+\frac{10}{53}i.

$$\frac{35}{53}+\frac{10}{53}i$$

Real Part

$$\frac{35}{53} = 0.660377358490566$$

Solution Steps

Multiply both numerator and denominator of \frac{5}{7-2i} by the complex conjugate of the denominator, 7+2i.

$$Re(\frac{5\left(7+2i\right)}{\left(7-2i\right)\left(7+2i\right)})$$

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

$$Re(\frac{5\left(7+2i\right)}{7^{2}-2^{2}i^{2}})$$

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

$$Re(\frac{5\left(7+2i\right)}{53})$$

Multiply 5 times 7+2i.

$$Re(\frac{5\times 7+5\times \left(2i\right)}{53})$$

Do the multiplications in 5\times 7+5\times \left(2i\right).

$$Re(\frac{35+10i}{53})$$

Divide 35+10i by 53 to get \frac{35}{53}+\frac{10}{53}i.

$$Re(\frac{35}{53}+\frac{10}{53}i)$$

The real part of \frac{35}{53}+\frac{10}{53}i is \frac{35}{53}.

$$\frac{35}{53}$$