Solve 4i(2-i)(5+3i)

$$4 i ( 2 - i ) ( 5 + 3 i )$$

Evaluate

$$-4+52i$$

Solution Steps

Multiply 4i times 2-i.

$$\left(4i\times 2+4\left(-1\right)i^{2}\right)\left(5+3i\right)$$

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

$$\left(4i\times 2+4\left(-1\right)\left(-1\right)\right)\left(5+3i\right)$$

Do the multiplications. Reorder the terms.

$$\left(4+8i\right)\left(5+3i\right)$$

Multiply complex numbers 4+8i and 5+3i like you multiply binomials.

$$4\times 5+4\times \left(3i\right)+8i\times 5+8\times 3i^{2}$$

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

$$4\times 5+4\times \left(3i\right)+8i\times 5+8\times 3\left(-1\right)$$

Do the multiplications.

$$20+12i+40i-24$$

Combine the real and imaginary parts.

$$20-24+\left(12+40\right)i$$

Do the additions.

$$-4+52i$$

Real Part

$$-4$$

Solution Steps

Multiply 4i times 2-i.

$$Re(\left(4i\times 2+4\left(-1\right)i^{2}\right)\left(5+3i\right))$$

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

$$Re(\left(4i\times 2+4\left(-1\right)\left(-1\right)\right)\left(5+3i\right))$$

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

$$Re(\left(4+8i\right)\left(5+3i\right))$$

Multiply complex numbers 4+8i and 5+3i like you multiply binomials.

$$Re(4\times 5+4\times \left(3i\right)+8i\times 5+8\times 3i^{2})$$

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

$$Re(4\times 5+4\times \left(3i\right)+8i\times 5+8\times 3\left(-1\right))$$

Do the multiplications in 4\times 5+4\times \left(3i\right)+8i\times 5+8\times 3\left(-1\right).

$$Re(20+12i+40i-24)$$

Combine the real and imaginary parts in 20+12i+40i-24.

$$Re(20-24+\left(12+40\right)i)$$

Do the additions in 20-24+\left(12+40\right)i.

$$Re(-4+52i)$$

The real part of -4+52i is -4.

$$-4$$