Solve 1/2(8-10i)[2(3-i)]

$$\frac { 1 } { 2 } ( 8 - 10 i ) [ 2 ( 3 - i ) ]$$

Evaluate

$$14-38i$$

Solution Steps

Multiply \frac{1}{2} times 8-10i.

$$\left(\frac{1}{2}\times 8+\frac{1}{2}\times \left(-10i\right)\right)\times 2\left(3-i\right)$$

Do the multiplications.

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

Multiply 4-5i times 2.

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

Do the multiplications.

$$\left(8-10i\right)\left(3-i\right)$$

Multiply complex numbers 8-10i and 3-i like you multiply binomials.

$$8\times 3+8\left(-i\right)-10i\times 3-10\left(-1\right)i^{2}$$

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

$$8\times 3+8\left(-i\right)-10i\times 3-10\left(-1\right)\left(-1\right)$$

Do the multiplications.

$$24-8i-30i-10$$

Combine the real and imaginary parts.

$$24-10+\left(-8-30\right)i$$

Do the additions.

$$14-38i$$

Real Part

$$14$$

Solution Steps

Multiply \frac{1}{2} times 8-10i.

$$Re(\left(\frac{1}{2}\times 8+\frac{1}{2}\times \left(-10i\right)\right)\times 2\left(3-i\right))$$

Do the multiplications in \frac{1}{2}\times 8+\frac{1}{2}\times \left(-10i\right).

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

Multiply 4-5i times 2.

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

Do the multiplications in 4\times 2-5i\times 2.

$$Re(\left(8-10i\right)\left(3-i\right))$$

Multiply complex numbers 8-10i and 3-i like you multiply binomials.

$$Re(8\times 3+8\left(-i\right)-10i\times 3-10\left(-1\right)i^{2})$$

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

$$Re(8\times 3+8\left(-i\right)-10i\times 3-10\left(-1\right)\left(-1\right))$$

Do the multiplications in 8\times 3+8\left(-i\right)-10i\times 3-10\left(-1\right)\left(-1\right).

$$Re(24-8i-30i-10)$$

Combine the real and imaginary parts in 24-8i-30i-10.

$$Re(24-10+\left(-8-30\right)i)$$

Do the additions in 24-10+\left(-8-30\right)i.

$$Re(14-38i)$$

The real part of 14-38i is 14.

$$14$$