Solve 1/(4+3i)(3-2i)

$$\frac { 1 } { ( 4 + 3 i ) ( 3 - 2 i ) }$$

Evaluate

$$\frac{18}{325}-\frac{1}{325}i\approx 0.055384615-0.003076923i$$

Solution Steps

Multiply complex numbers 4+3i and 3-2i like you multiply binomials.

$$\frac{1}{4\times 3+4\times \left(-2i\right)+3i\times 3+3\left(-2\right)i^{2}}$$

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

$$\frac{1}{4\times 3+4\times \left(-2i\right)+3i\times 3+3\left(-2\right)\left(-1\right)}$$

Do the multiplications in 4\times 3+4\times \left(-2i\right)+3i\times 3+3\left(-2\right)\left(-1\right).

$$\frac{1}{12-8i+9i+6}$$

Combine the real and imaginary parts in 12-8i+9i+6.

$$\frac{1}{12+6+\left(-8+9\right)i}$$

Do the additions in 12+6+\left(-8+9\right)i.

$$\frac{1}{18+i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 18-i.

$$\frac{1\left(18-i\right)}{\left(18+i\right)\left(18-i\right)}$$

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

$$\frac{1\left(18-i\right)}{18^{2}-i^{2}}$$

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

$$\frac{1\left(18-i\right)}{325}$$

Multiply 1 and 18-i to get 18-i.

$$\frac{18-i}{325}$$

Divide 18-i by 325 to get \frac{18}{325}-\frac{1}{325}i.

$$\frac{18}{325}-\frac{1}{325}i$$

Real Part

$$\frac{18}{325} = 0.055384615384615386$$

Solution Steps

Multiply complex numbers 4+3i and 3-2i like you multiply binomials.

$$Re(\frac{1}{4\times 3+4\times \left(-2i\right)+3i\times 3+3\left(-2\right)i^{2}})$$

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

$$Re(\frac{1}{4\times 3+4\times \left(-2i\right)+3i\times 3+3\left(-2\right)\left(-1\right)})$$

Do the multiplications in 4\times 3+4\times \left(-2i\right)+3i\times 3+3\left(-2\right)\left(-1\right).

$$Re(\frac{1}{12-8i+9i+6})$$

Combine the real and imaginary parts in 12-8i+9i+6.

$$Re(\frac{1}{12+6+\left(-8+9\right)i})$$

Do the additions in 12+6+\left(-8+9\right)i.

$$Re(\frac{1}{18+i})$$

Multiply both numerator and denominator of \frac{1}{18+i} by the complex conjugate of the denominator, 18-i.

$$Re(\frac{1\left(18-i\right)}{\left(18+i\right)\left(18-i\right)})$$

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

$$Re(\frac{1\left(18-i\right)}{18^{2}-i^{2}})$$

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

$$Re(\frac{1\left(18-i\right)}{325})$$

Multiply 1 and 18-i to get 18-i.

$$Re(\frac{18-i}{325})$$

Divide 18-i by 325 to get \frac{18}{325}-\frac{1}{325}i.

$$Re(\frac{18}{325}-\frac{1}{325}i)$$

The real part of \frac{18}{325}-\frac{1}{325}i is \frac{18}{325}.

$$\frac{18}{325}$$