Solve 8+3i/-9+4iquad(-9-7)

$$\frac { 8 + 3 i } { - 9 + 4 i } \quad ( - 9 - 7 )$$

Evaluate

$$\frac{960}{97}+\frac{944}{97}i\approx 9.896907216+9.731958763i$$

Solution Steps

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

$$\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9+4i\right)\left(-9-4i\right)}\left(-9-7\right)$$

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

$$\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9\right)^{2}-4^{2}i^{2}}\left(-9-7\right)$$

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

$$\frac{\left(8+3i\right)\left(-9-4i\right)}{97}\left(-9-7\right)$$

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

$$\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)i^{2}}{97}\left(-9-7\right)$$

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

$$\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)\left(-1\right)}{97}\left(-9-7\right)$$

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

$$\frac{-72-32i-27i+12}{97}\left(-9-7\right)$$

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

$$\frac{-72+12+\left(-32-27\right)i}{97}\left(-9-7\right)$$

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

$$\frac{-60-59i}{97}\left(-9-7\right)$$

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

$$\left(-\frac{60}{97}-\frac{59}{97}i\right)\left(-9-7\right)$$

Subtract 7 from -9 to get -16.

$$\left(-\frac{60}{97}-\frac{59}{97}i\right)\left(-16\right)$$

Multiply -\frac{60}{97}-\frac{59}{97}i times -16.

$$-\frac{60}{97}\left(-16\right)-\frac{59}{97}i\left(-16\right)$$

Do the multiplications.

$$\frac{960}{97}+\frac{944}{97}i$$

Real Part

$$\frac{960}{97} = 9\frac{87}{97} = 9.896907216494846$$

Solution Steps

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

$$Re(\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9+4i\right)\left(-9-4i\right)}\left(-9-7\right))$$

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

$$Re(\frac{\left(8+3i\right)\left(-9-4i\right)}{\left(-9\right)^{2}-4^{2}i^{2}}\left(-9-7\right))$$

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

$$Re(\frac{\left(8+3i\right)\left(-9-4i\right)}{97}\left(-9-7\right))$$

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

$$Re(\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)i^{2}}{97}\left(-9-7\right))$$

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

$$Re(\frac{8\left(-9\right)+8\times \left(-4i\right)+3i\left(-9\right)+3\left(-4\right)\left(-1\right)}{97}\left(-9-7\right))$$

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

$$Re(\frac{-72-32i-27i+12}{97}\left(-9-7\right))$$

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

$$Re(\frac{-72+12+\left(-32-27\right)i}{97}\left(-9-7\right))$$

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

$$Re(\frac{-60-59i}{97}\left(-9-7\right))$$

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

$$Re(\left(-\frac{60}{97}-\frac{59}{97}i\right)\left(-9-7\right))$$

Subtract 7 from -9 to get -16.

$$Re(\left(-\frac{60}{97}-\frac{59}{97}i\right)\left(-16\right))$$

Multiply -\frac{60}{97}-\frac{59}{97}i times -16.

$$Re(-\frac{60}{97}\left(-16\right)-\frac{59}{97}i\left(-16\right))$$

Do the multiplications in -\frac{60}{97}\left(-16\right)-\frac{59}{97}i\left(-16\right).

$$Re(\frac{960}{97}+\frac{944}{97}i)$$

The real part of \frac{960}{97}+\frac{944}{97}i is \frac{960}{97}.

$$\frac{960}{97}$$