Solve (3div2i)*(4+3i)(3+2i)

$$( 3 \div 2 i ) \cdot ( 4 + 3 i ) ( 3 + 2 i )$$

Evaluate

$$-\frac{51}{2}+9i=-25.5+9i$$

Solution Steps

Multiply \frac{3}{2}i times 4+3i.

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

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

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

Do the multiplications. Reorder the terms.

$$\left(-\frac{9}{2}+6i\right)\left(3+2i\right)$$

Multiply complex numbers -\frac{9}{2}+6i and 3+2i like you multiply binomials.

$$-\frac{9}{2}\times 3-\frac{9}{2}\times \left(2i\right)+6i\times 3+6\times 2i^{2}$$

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

$$-\frac{9}{2}\times 3-\frac{9}{2}\times \left(2i\right)+6i\times 3+6\times 2\left(-1\right)$$

Do the multiplications.

$$-\frac{27}{2}-9i+18i-12$$

Combine the real and imaginary parts.

$$-\frac{27}{2}-12+\left(-9+18\right)i$$

Do the additions.

$$-\frac{51}{2}+9i$$

Real Part

$$-\frac{51}{2} = -25\frac{1}{2} = -25.5$$

Solution Steps

Multiply \frac{3}{2}i times 4+3i.

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

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

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

Do the multiplications in \frac{3}{2}i\times 4+\frac{3}{2}\times 3\left(-1\right). Reorder the terms.

$$Re(\left(-\frac{9}{2}+6i\right)\left(3+2i\right))$$

Multiply complex numbers -\frac{9}{2}+6i and 3+2i like you multiply binomials.

$$Re(-\frac{9}{2}\times 3-\frac{9}{2}\times \left(2i\right)+6i\times 3+6\times 2i^{2})$$

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

$$Re(-\frac{9}{2}\times 3-\frac{9}{2}\times \left(2i\right)+6i\times 3+6\times 2\left(-1\right))$$

Do the multiplications in -\frac{9}{2}\times 3-\frac{9}{2}\times \left(2i\right)+6i\times 3+6\times 2\left(-1\right).

$$Re(-\frac{27}{2}-9i+18i-12)$$

Combine the real and imaginary parts in -\frac{27}{2}-9i+18i-12.

$$Re(-\frac{27}{2}-12+\left(-9+18\right)i)$$

Do the additions in -\frac{27}{2}-12+\left(-9+18\right)i.

$$Re(-\frac{51}{2}+9i)$$

The real part of -\frac{51}{2}+9i is -\frac{51}{2}.

$$-\frac{51}{2}$$