Solve 4-3i/3+4i

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

Evaluate

$$-i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$\frac{12-16i-9i-12}{25}$$

Combine the real and imaginary parts in 12-16i-9i-12.

$$\frac{12-12+\left(-16-9\right)i}{25}$$

Do the additions in 12-12+\left(-16-9\right)i.

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

Divide -25i by 25 to get -i.

$$-i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

$$Re(\frac{12-16i-9i-12}{25})$$

Combine the real and imaginary parts in 12-16i-9i-12.

$$Re(\frac{12-12+\left(-16-9\right)i}{25})$$

Do the additions in 12-12+\left(-16-9\right)i.

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

Divide -25i by 25 to get -i.

$$Re(-i)$$

The real part of -i is 0.

$$0$$