Solve 3/5-6i

$$\frac { 3 } { 5 - 6 i }$$

Evaluate

$$\frac{15}{61}+\frac{18}{61}i\approx 0.245901639+0.295081967i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 5+6i.

$$\frac{3\left(5+6i\right)}{\left(5-6i\right)\left(5+6i\right)}$$

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

$$\frac{3\left(5+6i\right)}{5^{2}-6^{2}i^{2}}$$

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

$$\frac{3\left(5+6i\right)}{61}$$

Multiply 3 times 5+6i.

$$\frac{3\times 5+3\times \left(6i\right)}{61}$$

Do the multiplications in 3\times 5+3\times \left(6i\right).

$$\frac{15+18i}{61}$$

Divide 15+18i by 61 to get \frac{15}{61}+\frac{18}{61}i.

$$\frac{15}{61}+\frac{18}{61}i$$

Real Part

$$\frac{15}{61} = 0.2459016393442623$$

Solution Steps

Multiply both numerator and denominator of \frac{3}{5-6i} by the complex conjugate of the denominator, 5+6i.

$$Re(\frac{3\left(5+6i\right)}{\left(5-6i\right)\left(5+6i\right)})$$

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

$$Re(\frac{3\left(5+6i\right)}{5^{2}-6^{2}i^{2}})$$

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

$$Re(\frac{3\left(5+6i\right)}{61})$$

Multiply 3 times 5+6i.

$$Re(\frac{3\times 5+3\times \left(6i\right)}{61})$$

Do the multiplications in 3\times 5+3\times \left(6i\right).

$$Re(\frac{15+18i}{61})$$

Divide 15+18i by 61 to get \frac{15}{61}+\frac{18}{61}i.

$$Re(\frac{15}{61}+\frac{18}{61}i)$$

The real part of \frac{15}{61}+\frac{18}{61}i is \frac{15}{61}.

$$\frac{15}{61}$$