Solve 3-4i/-7+9i

$$\frac { 3 - 4 i } { - 7 + 9 i }$$

Evaluate

$$-\frac{57}{130}+\frac{1}{130}i\approx -0.438461538+0.007692308i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -7-9i.

$$\frac{\left(3-4i\right)\left(-7-9i\right)}{\left(-7+9i\right)\left(-7-9i\right)}$$

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

$$\frac{\left(3-4i\right)\left(-7-9i\right)}{\left(-7\right)^{2}-9^{2}i^{2}}$$

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

$$\frac{\left(3-4i\right)\left(-7-9i\right)}{130}$$

Multiply complex numbers 3-4i and -7-9i like you multiply binomials.

$$\frac{3\left(-7\right)+3\times \left(-9i\right)-4i\left(-7\right)-4\left(-9\right)i^{2}}{130}$$

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

$$\frac{3\left(-7\right)+3\times \left(-9i\right)-4i\left(-7\right)-4\left(-9\right)\left(-1\right)}{130}$$

Do the multiplications in 3\left(-7\right)+3\times \left(-9i\right)-4i\left(-7\right)-4\left(-9\right)\left(-1\right).

$$\frac{-21-27i+28i-36}{130}$$

Combine the real and imaginary parts in -21-27i+28i-36.

$$\frac{-21-36+\left(-27+28\right)i}{130}$$

Do the additions in -21-36+\left(-27+28\right)i.

$$\frac{-57+i}{130}$$

Divide -57+i by 130 to get -\frac{57}{130}+\frac{1}{130}i.

$$-\frac{57}{130}+\frac{1}{130}i$$

Real Part

$$-\frac{57}{130} = -0.43846153846153846$$

Solution Steps

Multiply both numerator and denominator of \frac{3-4i}{-7+9i} by the complex conjugate of the denominator, -7-9i.

$$Re(\frac{\left(3-4i\right)\left(-7-9i\right)}{\left(-7+9i\right)\left(-7-9i\right)})$$

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

$$Re(\frac{\left(3-4i\right)\left(-7-9i\right)}{\left(-7\right)^{2}-9^{2}i^{2}})$$

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

$$Re(\frac{\left(3-4i\right)\left(-7-9i\right)}{130})$$

Multiply complex numbers 3-4i and -7-9i like you multiply binomials.

$$Re(\frac{3\left(-7\right)+3\times \left(-9i\right)-4i\left(-7\right)-4\left(-9\right)i^{2}}{130})$$

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

$$Re(\frac{3\left(-7\right)+3\times \left(-9i\right)-4i\left(-7\right)-4\left(-9\right)\left(-1\right)}{130})$$

Do the multiplications in 3\left(-7\right)+3\times \left(-9i\right)-4i\left(-7\right)-4\left(-9\right)\left(-1\right).

$$Re(\frac{-21-27i+28i-36}{130})$$

Combine the real and imaginary parts in -21-27i+28i-36.

$$Re(\frac{-21-36+\left(-27+28\right)i}{130})$$

Do the additions in -21-36+\left(-27+28\right)i.

$$Re(\frac{-57+i}{130})$$

Divide -57+i by 130 to get -\frac{57}{130}+\frac{1}{130}i.

$$Re(-\frac{57}{130}+\frac{1}{130}i)$$

The real part of -\frac{57}{130}+\frac{1}{130}i is -\frac{57}{130}.

$$-\frac{57}{130}$$