Solve 3+2i*(1+6i)

$$3 + 2 i \cdot ( 1 + 6 i )$$

Evaluate

$$-9+2i$$

Solution Steps

Multiply 2i times 1+6i.

$$3+2i\times 1+2\times 6i^{2}$$

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

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

Do the multiplications in 2i\times 1+2\times 6\left(-1\right). Reorder the terms.

$$3+\left(-12+2i\right)$$

Combine the real and imaginary parts.

$$3-12+2i$$

Add 3 to -12.

$$-9+2i$$

Real Part

$$-9$$

Solution Steps

Multiply 2i times 1+6i.

$$Re(3+2i\times 1+2\times 6i^{2})$$

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

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

Do the multiplications in 2i\times 1+2\times 6\left(-1\right). Reorder the terms.

$$Re(3+\left(-12+2i\right))$$

Combine the real and imaginary parts in 3-12+2i.

$$Re(3-12+2i)$$

Add 3 to -12.

$$Re(-9+2i)$$

The real part of -9+2i is -9.

$$-9$$