Solve 16-9i/7+24i

$$\frac{ 16-9i }{ 7+24i }$$

Evaluate

$$-\frac{104}{625}-\frac{447}{625}i=-0.1664-0.7152i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 7-24i.

$$\frac{\left(16-9i\right)\left(7-24i\right)}{\left(7+24i\right)\left(7-24i\right)}$$

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

$$\frac{\left(16-9i\right)\left(7-24i\right)}{7^{2}-24^{2}i^{2}}$$

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

$$\frac{\left(16-9i\right)\left(7-24i\right)}{625}$$

Multiply complex numbers 16-9i and 7-24i like you multiply binomials.

$$\frac{16\times 7+16\times \left(-24i\right)-9i\times 7-9\left(-24\right)i^{2}}{625}$$

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

$$\frac{16\times 7+16\times \left(-24i\right)-9i\times 7-9\left(-24\right)\left(-1\right)}{625}$$

Do the multiplications in 16\times 7+16\times \left(-24i\right)-9i\times 7-9\left(-24\right)\left(-1\right).

$$\frac{112-384i-63i-216}{625}$$

Combine the real and imaginary parts in 112-384i-63i-216.

$$\frac{112-216+\left(-384-63\right)i}{625}$$

Do the additions in 112-216+\left(-384-63\right)i.

$$\frac{-104-447i}{625}$$

Divide -104-447i by 625 to get -\frac{104}{625}-\frac{447}{625}i.

$$-\frac{104}{625}-\frac{447}{625}i$$

Real Part

$$-\frac{104}{625} = -0.1664$$

Solution Steps

Multiply both numerator and denominator of \frac{16-9i}{7+24i} by the complex conjugate of the denominator, 7-24i.

$$Re(\frac{\left(16-9i\right)\left(7-24i\right)}{\left(7+24i\right)\left(7-24i\right)})$$

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

$$Re(\frac{\left(16-9i\right)\left(7-24i\right)}{7^{2}-24^{2}i^{2}})$$

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

$$Re(\frac{\left(16-9i\right)\left(7-24i\right)}{625})$$

Multiply complex numbers 16-9i and 7-24i like you multiply binomials.

$$Re(\frac{16\times 7+16\times \left(-24i\right)-9i\times 7-9\left(-24\right)i^{2}}{625})$$

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

$$Re(\frac{16\times 7+16\times \left(-24i\right)-9i\times 7-9\left(-24\right)\left(-1\right)}{625})$$

Do the multiplications in 16\times 7+16\times \left(-24i\right)-9i\times 7-9\left(-24\right)\left(-1\right).

$$Re(\frac{112-384i-63i-216}{625})$$

Combine the real and imaginary parts in 112-384i-63i-216.

$$Re(\frac{112-216+\left(-384-63\right)i}{625})$$

Do the additions in 112-216+\left(-384-63\right)i.

$$Re(\frac{-104-447i}{625})$$

Divide -104-447i by 625 to get -\frac{104}{625}-\frac{447}{625}i.

$$Re(-\frac{104}{625}-\frac{447}{625}i)$$

The real part of -\frac{104}{625}-\frac{447}{625}i is -\frac{104}{625}.

$$-\frac{104}{625}$$