Solve (6+7i)div(-3+4i)

$$( 6 + 7 i ) \div ( - 3 + 4 i )$$

Evaluate

$$\frac{2}{5}-\frac{9}{5}i=0.4-1.8i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -3-4i.

$$\frac{\left(6+7i\right)\left(-3-4i\right)}{\left(-3+4i\right)\left(-3-4i\right)}$$

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

$$\frac{\left(6+7i\right)\left(-3-4i\right)}{\left(-3\right)^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(6+7i\right)\left(-3-4i\right)}{25}$$

Multiply complex numbers 6+7i and -3-4i like you multiply binomials.

$$\frac{6\left(-3\right)+6\times \left(-4i\right)+7i\left(-3\right)+7\left(-4\right)i^{2}}{25}$$

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

$$\frac{6\left(-3\right)+6\times \left(-4i\right)+7i\left(-3\right)+7\left(-4\right)\left(-1\right)}{25}$$

Do the multiplications in 6\left(-3\right)+6\times \left(-4i\right)+7i\left(-3\right)+7\left(-4\right)\left(-1\right).

$$\frac{-18-24i-21i+28}{25}$$

Combine the real and imaginary parts in -18-24i-21i+28.

$$\frac{-18+28+\left(-24-21\right)i}{25}$$

Do the additions in -18+28+\left(-24-21\right)i.

$$\frac{10-45i}{25}$$

Divide 10-45i by 25 to get \frac{2}{5}-\frac{9}{5}i.

$$\frac{2}{5}-\frac{9}{5}i$$

Real Part

$$\frac{2}{5} = 0.4$$

Solution Steps

Multiply both numerator and denominator of \frac{6+7i}{-3+4i} by the complex conjugate of the denominator, -3-4i.

$$Re(\frac{\left(6+7i\right)\left(-3-4i\right)}{\left(-3+4i\right)\left(-3-4i\right)})$$

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

$$Re(\frac{\left(6+7i\right)\left(-3-4i\right)}{\left(-3\right)^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(6+7i\right)\left(-3-4i\right)}{25})$$

Multiply complex numbers 6+7i and -3-4i like you multiply binomials.

$$Re(\frac{6\left(-3\right)+6\times \left(-4i\right)+7i\left(-3\right)+7\left(-4\right)i^{2}}{25})$$

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

$$Re(\frac{6\left(-3\right)+6\times \left(-4i\right)+7i\left(-3\right)+7\left(-4\right)\left(-1\right)}{25})$$

Do the multiplications in 6\left(-3\right)+6\times \left(-4i\right)+7i\left(-3\right)+7\left(-4\right)\left(-1\right).

$$Re(\frac{-18-24i-21i+28}{25})$$

Combine the real and imaginary parts in -18-24i-21i+28.

$$Re(\frac{-18+28+\left(-24-21\right)i}{25})$$

Do the additions in -18+28+\left(-24-21\right)i.

$$Re(\frac{10-45i}{25})$$

Divide 10-45i by 25 to get \frac{2}{5}-\frac{9}{5}i.

$$Re(\frac{2}{5}-\frac{9}{5}i)$$

The real part of \frac{2}{5}-\frac{9}{5}i is \frac{2}{5}.

$$\frac{2}{5}$$