Solve 3+4i/4-3i

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

Evaluate

$$i$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Do the additions in 12-12+\left(9+16\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{3+4i}{4-3i} by the complex conjugate of the denominator, 4+3i.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Divide 25i by 25 to get i.

$$Re(i)$$

The real part of i is 0.

$$0$$