Solve (16+4i)-(8-6i)/2i

$$\frac { ( 16 + 4 i ) - ( 8 - 6 i ) } { 2 i }$$

Evaluate

$$5-4i$$

Solution Steps

Subtract 8-6i from 16+4i by subtracting corresponding real and imaginary parts.

$$\frac{16-8+\left(4-\left(-6\right)\right)i}{2i}$$

Subtract 8 from 16. Subtract -6 from 4.

$$\frac{8+10i}{2i}$$

Multiply both numerator and denominator by imaginary unit i.

$$\frac{\left(8+10i\right)i}{2i^{2}}$$

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

$$\frac{\left(8+10i\right)i}{-2}$$

Multiply 8+10i times i.

$$\frac{8i+10i^{2}}{-2}$$

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

$$\frac{8i+10\left(-1\right)}{-2}$$

Do the multiplications in 8i+10\left(-1\right). Reorder the terms.

$$\frac{-10+8i}{-2}$$

Divide -10+8i by -2 to get 5-4i.

$$5-4i$$

Real Part

$$5$$

Solution Steps

Subtract 8-6i from 16+4i by subtracting corresponding real and imaginary parts.

$$Re(\frac{16-8+\left(4-\left(-6\right)\right)i}{2i})$$

Subtract 8 from 16. Subtract -6 from 4.

$$Re(\frac{8+10i}{2i})$$

Multiply both numerator and denominator of \frac{8+10i}{2i} by imaginary unit i.

$$Re(\frac{\left(8+10i\right)i}{2i^{2}})$$

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

$$Re(\frac{\left(8+10i\right)i}{-2})$$

Multiply 8+10i times i.

$$Re(\frac{8i+10i^{2}}{-2})$$

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

$$Re(\frac{8i+10\left(-1\right)}{-2})$$

Do the multiplications in 8i+10\left(-1\right). Reorder the terms.

$$Re(\frac{-10+8i}{-2})$$

Divide -10+8i by -2 to get 5-4i.

$$Re(5-4i)$$

The real part of 5-4i is 5.

$$5$$