Solve 3/3-7i

$$\frac { 3 } { 3 - 7 i }$$

Evaluate

$$\frac{9}{58}+\frac{21}{58}i\approx 0.155172414+0.362068966i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 3+7i.

$$\frac{3\left(3+7i\right)}{\left(3-7i\right)\left(3+7i\right)}$$

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

$$\frac{3\left(3+7i\right)}{3^{2}-7^{2}i^{2}}$$

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

$$\frac{3\left(3+7i\right)}{58}$$

Multiply 3 times 3+7i.

$$\frac{3\times 3+3\times \left(7i\right)}{58}$$

Do the multiplications in 3\times 3+3\times \left(7i\right).

$$\frac{9+21i}{58}$$

Divide 9+21i by 58 to get \frac{9}{58}+\frac{21}{58}i.

$$\frac{9}{58}+\frac{21}{58}i$$

Real Part

$$\frac{9}{58} = 0.15517241379310345$$

Solution Steps

Multiply both numerator and denominator of \frac{3}{3-7i} by the complex conjugate of the denominator, 3+7i.

$$Re(\frac{3\left(3+7i\right)}{\left(3-7i\right)\left(3+7i\right)})$$

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

$$Re(\frac{3\left(3+7i\right)}{3^{2}-7^{2}i^{2}})$$

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

$$Re(\frac{3\left(3+7i\right)}{58})$$

Multiply 3 times 3+7i.

$$Re(\frac{3\times 3+3\times \left(7i\right)}{58})$$

Do the multiplications in 3\times 3+3\times \left(7i\right).

$$Re(\frac{9+21i}{58})$$

Divide 9+21i by 58 to get \frac{9}{58}+\frac{21}{58}i.

$$Re(\frac{9}{58}+\frac{21}{58}i)$$

The real part of \frac{9}{58}+\frac{21}{58}i is \frac{9}{58}.

$$\frac{9}{58}$$