Solve (-4+3i)+2(8+1i)-i(2+1i)

$$( - 4 + 3 i ) + 2 ( 8 + 1 i ) - i ( 2 + 1 i )$$

Evaluate

$$13+3i$$

Solution Steps

Multiply 2 times 8+i.

$$-4+3i+2\times 8+2i-i\left(2+i\right)$$

Do the multiplications in 2\times 8+2i.

$$-4+3i+\left(16+2i\right)-i\left(2+i\right)$$

Combine the real and imaginary parts in -4+3i+16+2i.

$$-4+16+\left(3+2\right)i-i\left(2+i\right)$$

Do the additions in -4+16+\left(3+2\right)i.

$$12+5i-i\left(2+i\right)$$

Multiply i times 2+i.

$$12+5i-\left(2i+i^{2}\right)$$

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

$$12+5i-\left(2i-1\right)$$

Reorder the terms.

$$12+5i-\left(-1+2i\right)$$

Subtract -1+2i from 12+5i by subtracting corresponding real and imaginary parts.

$$12-\left(-1\right)+\left(5-2\right)i$$

Subtract -1 from 12. Subtract 2 from 5.

$$13+3i$$

Real Part

$$13$$

Solution Steps

Multiply 2 times 8+i.

$$Re(-4+3i+2\times 8+2i-i\left(2+i\right))$$

Do the multiplications in 2\times 8+2i.

$$Re(-4+3i+\left(16+2i\right)-i\left(2+i\right))$$

Combine the real and imaginary parts in -4+3i+16+2i.

$$Re(-4+16+\left(3+2\right)i-i\left(2+i\right))$$

Do the additions in -4+16+\left(3+2\right)i.

$$Re(12+5i-i\left(2+i\right))$$

Multiply i times 2+i.

$$Re(12+5i-\left(2i+i^{2}\right))$$

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

$$Re(12+5i-\left(2i-1\right))$$

Reorder the terms.

$$Re(12+5i-\left(-1+2i\right))$$

Subtract -1+2i from 12+5i by subtracting corresponding real and imaginary parts.

$$Re(12-\left(-1\right)+\left(5-2\right)i)$$

Subtract -1 from 12. Subtract 2 from 5.

$$Re(13+3i)$$

The real part of 13+3i is 13.

$$13$$