Solve (2+4i)*(3-2i)Rightarrow(

$$( 2 + 4 i ) \cdot ( 3 - 2 i ) \Rightarrow ($$

Evaluate

$$14+8i$$

Solution Steps

Multiply complex numbers 2+4i and 3-2i like you multiply binomials.

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

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

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

Do the multiplications.

$$6-4i+12i+8$$

Combine the real and imaginary parts.

$$6+8+\left(-4+12\right)i$$

Do the additions.

$$14+8i$$

Real Part

$$14$$

Solution Steps

Multiply complex numbers 2+4i and 3-2i like you multiply binomials.

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

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

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

Do the multiplications in 2\times 3+2\times \left(-2i\right)+4i\times 3+4\left(-2\right)\left(-1\right).

$$Re(6-4i+12i+8)$$

Combine the real and imaginary parts in 6-4i+12i+8.

$$Re(6+8+\left(-4+12\right)i)$$

Do the additions in 6+8+\left(-4+12\right)i.

$$Re(14+8i)$$

The real part of 14+8i is 14.

$$14$$