Solve 2+5i-2i(3+2i)

$$2 + 5 i - 2 i ( 3 + 2 i )$$

Evaluate

$$6-i$$

Solution Steps

Multiply 2i times 3+2i.

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

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

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

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

$$2+5i-\left(-4+6i\right)$$

Subtract -4+6i from 2+5i by subtracting corresponding real and imaginary parts.

$$2-\left(-4\right)+\left(5-6\right)i$$

Subtract -4 from 2. Subtract 6 from 5.

$$6-i$$

Real Part

$$6$$

Solution Steps

Multiply 2i times 3+2i.

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

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

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

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

$$Re(2+5i-\left(-4+6i\right))$$

Subtract -4+6i from 2+5i by subtracting corresponding real and imaginary parts.

$$Re(2-\left(-4\right)+\left(5-6\right)i)$$

Subtract -4 from 2. Subtract 6 from 5.

$$Re(6-i)$$

The real part of 6-i is 6.

$$6$$