Solve (-6+2i)*(1/2+i)

$$( - 6 + 2 i ) \cdot ( \frac { 1 } { 2 } + i )$$

Evaluate

$$-5-5i$$

Solution Steps

Multiply complex numbers -6+2i and \frac{1}{2}+i like you multiply binomials.

$$-6\times \frac{1}{2}-6i+2i\times \frac{1}{2}+2i^{2}$$

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

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

Do the multiplications.

$$-3-6i+i-2$$

Combine the real and imaginary parts.

$$-3-2+\left(-6+1\right)i$$

Do the additions.

$$-5-5i$$

Real Part

$$-5$$

Solution Steps

Multiply complex numbers -6+2i and \frac{1}{2}+i like you multiply binomials.

$$Re(-6\times \frac{1}{2}-6i+2i\times \frac{1}{2}+2i^{2})$$

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

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

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

$$Re(-3-6i+i-2)$$

Combine the real and imaginary parts in -3-6i+i-2.

$$Re(-3-2+\left(-6+1\right)i)$$

Do the additions in -3-2+\left(-6+1\right)i.

$$Re(-5-5i)$$

The real part of -5-5i is -5.

$$-5$$