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

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

Evaluate

$$1-18i$$

Solution Steps

Divide i by i to get 1.

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

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

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

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

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

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

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

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

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

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

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

Multiply both numerator and denominator of \frac{18+i}{i} by imaginary unit i.

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

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

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

Multiply 18+i times i.

$$\frac{18i+i^{2}}{-1}\times 1$$

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

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

Reorder the terms.

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

Divide -1+18i by -1 to get 1-18i.

$$\left(1-18i\right)\times 1$$

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

$$1-18i$$

Real Part

$$1$$

Solution Steps

Divide i by i to get 1.

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

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

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

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

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

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

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

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

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

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

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

Multiply both numerator and denominator of \frac{18+i}{i} by imaginary unit i.

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

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

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

Multiply 18+i times i.

$$Re(\frac{18i+i^{2}}{-1}\times 1)$$

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

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

Reorder the terms.

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

Divide -1+18i by -1 to get 1-18i.

$$Re(\left(1-18i\right)\times 1)$$

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

$$Re(1-18i)$$

The real part of 1-18i is 1.

$$1$$