Solve (5+2i)(-2+8i)+(3-i)(4+6i)=

$$( 5 + 2 i ) ( - 2 + 8 i ) + ( 3 - i ) ( 4 + 6 i ) =$$

Evaluate

$$-8+50i$$

Solution Steps

Multiply complex numbers 5+2i and -2+8i like you multiply binomials.

$$5\left(-2\right)+5\times \left(8i\right)+2i\left(-2\right)+2\times 8i^{2}+\left(3-i\right)\left(4+6i\right)$$

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

$$5\left(-2\right)+5\times \left(8i\right)+2i\left(-2\right)+2\times 8\left(-1\right)+\left(3-i\right)\left(4+6i\right)$$

Do the multiplications in 5\left(-2\right)+5\times \left(8i\right)+2i\left(-2\right)+2\times 8\left(-1\right).

$$-10+40i-4i-16+\left(3-i\right)\left(4+6i\right)$$

Combine the real and imaginary parts in -10+40i-4i-16.

$$-10-16+\left(40-4\right)i+\left(3-i\right)\left(4+6i\right)$$

Do the additions in -10-16+\left(40-4\right)i.

$$-26+36i+\left(3-i\right)\left(4+6i\right)$$

Multiply complex numbers 3-i and 4+6i like you multiply binomials.

$$-26+36i+3\times 4+3\times \left(6i\right)-i\times 4-6i^{2}$$

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

$$-26+36i+3\times 4+3\times \left(6i\right)-i\times 4-6\left(-1\right)$$

Do the multiplications in 3\times 4+3\times \left(6i\right)-i\times 4-6\left(-1\right).

$$-26+36i+12+18i-4i+6$$

Combine the real and imaginary parts in 12+18i-4i+6.

$$-26+36i+12+6+\left(18-4\right)i$$

Do the additions in 12+6+\left(18-4\right)i.

$$-26+36i+\left(18+14i\right)$$

Combine the real and imaginary parts.

$$-26+18+\left(36+14\right)i$$

Do the additions.

$$-8+50i$$

Real Part

$$-8$$

Solution Steps

Multiply complex numbers 5+2i and -2+8i like you multiply binomials.

$$Re(5\left(-2\right)+5\times \left(8i\right)+2i\left(-2\right)+2\times 8i^{2}+\left(3-i\right)\left(4+6i\right))$$

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

$$Re(5\left(-2\right)+5\times \left(8i\right)+2i\left(-2\right)+2\times 8\left(-1\right)+\left(3-i\right)\left(4+6i\right))$$

Do the multiplications in 5\left(-2\right)+5\times \left(8i\right)+2i\left(-2\right)+2\times 8\left(-1\right).

$$Re(-10+40i-4i-16+\left(3-i\right)\left(4+6i\right))$$

Combine the real and imaginary parts in -10+40i-4i-16.

$$Re(-10-16+\left(40-4\right)i+\left(3-i\right)\left(4+6i\right))$$

Do the additions in -10-16+\left(40-4\right)i.

$$Re(-26+36i+\left(3-i\right)\left(4+6i\right))$$

Multiply complex numbers 3-i and 4+6i like you multiply binomials.

$$Re(-26+36i+3\times 4+3\times \left(6i\right)-i\times 4-6i^{2})$$

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

$$Re(-26+36i+3\times 4+3\times \left(6i\right)-i\times 4-6\left(-1\right))$$

Do the multiplications in 3\times 4+3\times \left(6i\right)-i\times 4-6\left(-1\right).

$$Re(-26+36i+12+18i-4i+6)$$

Combine the real and imaginary parts in 12+18i-4i+6.

$$Re(-26+36i+12+6+\left(18-4\right)i)$$

Do the additions in 12+6+\left(18-4\right)i.

$$Re(-26+36i+\left(18+14i\right))$$

Combine the real and imaginary parts in -26+36i+18+14i.

$$Re(-26+18+\left(36+14\right)i)$$

Do the additions in -26+18+\left(36+14\right)i.

$$Re(-8+50i)$$

The real part of -8+50i is -8.

$$-8$$