Solve 8+3i/-9+4i

$$\frac{ 8+3i }{ -9+4i }$$

Evaluate

$$-\frac{60}{97}-\frac{59}{97}i\approx -0.618556701-0.608247423i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, -9-4i.

$$\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9+4i\right)\left(-9-4i\right)}$$

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

$$\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9\right)^{2}-4^{2}i^{2}}$$

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

$$\frac{\left(8+3i\right)\left(-9-4i\right)}{97}$$

Multiply complex numbers 8+3i and -9-4i like you multiply binomials.

$$\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)i^{2}}{97}$$

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

$$\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)\left(-1\right)}{97}$$

Do the multiplications in 8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)\left(-1\right).

$$\frac{-72-32i-27i+12}{97}$$

Combine the real and imaginary parts in -72-32i-27i+12.

$$\frac{-72+12+\left(-32-27\right)i}{97}$$

Do the additions in -72+12+\left(-32-27\right)i.

$$\frac{-60-59i}{97}$$

Divide -60-59i by 97 to get -\frac{60}{97}-\frac{59}{97}i.

$$-\frac{60}{97}-\frac{59}{97}i$$

Real Part

$$-\frac{60}{97} = -0.6185567010309279$$

Solution Steps

Multiply both numerator and denominator of \frac{8+3i}{-9+4i} by the complex conjugate of the denominator, -9-4i.

$$Re(\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9+4i\right)\left(-9-4i\right)})$$

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

$$Re(\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9\right)^{2}-4^{2}i^{2}})$$

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

$$Re(\frac{\left(8+3i\right)\left(-9-4i\right)}{97})$$

Multiply complex numbers 8+3i and -9-4i like you multiply binomials.

$$Re(\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)i^{2}}{97})$$

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

$$Re(\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)\left(-1\right)}{97})$$

Do the multiplications in 8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)\left(-1\right).

$$Re(\frac{-72-32i-27i+12}{97})$$

Combine the real and imaginary parts in -72-32i-27i+12.

$$Re(\frac{-72+12+\left(-32-27\right)i}{97})$$

Do the additions in -72+12+\left(-32-27\right)i.

$$Re(\frac{-60-59i}{97})$$

Divide -60-59i by 97 to get -\frac{60}{97}-\frac{59}{97}i.

$$Re(-\frac{60}{97}-\frac{59}{97}i)$$

The real part of -\frac{60}{97}-\frac{59}{97}i is -\frac{60}{97}.

$$-\frac{60}{97}$$