Solve -1/3i[(24-41i)-2(3-4i)]

$$- \frac { 1 } { 3 } i [ ( 24 - 41 i ) - 2 ( 3 - 4 i ) ]$$

Evaluate

$$-11-6i$$

Solution Steps

Multiply 2 times 3-4i.

$$-\frac{1}{3}i\left(24-41i-\left(2\times 3+2\times \left(-4i\right)\right)\right)$$

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

$$-\frac{1}{3}i\left(24-41i-\left(6-8i\right)\right)$$

Subtract 6-8i from 24-41i by subtracting corresponding real and imaginary parts.

$$-\frac{1}{3}i\left(24-6+\left(-41-\left(-8\right)\right)i\right)$$

Subtract 6 from 24. Subtract -8 from -41.

$$-\frac{1}{3}i\left(18-33i\right)$$

Multiply -\frac{1}{3}i times 18-33i.

$$-\frac{1}{3}i\times 18-\frac{1}{3}\left(-33\right)i^{2}$$

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

$$-\frac{1}{3}i\times 18-\frac{1}{3}\left(-33\right)\left(-1\right)$$

Do the multiplications. Reorder the terms.

$$-11-6i$$

Real Part

$$-11$$

Solution Steps

Multiply 2 times 3-4i.

$$Re(-\frac{1}{3}i\left(24-41i-\left(2\times 3+2\times \left(-4i\right)\right)\right))$$

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

$$Re(-\frac{1}{3}i\left(24-41i-\left(6-8i\right)\right))$$

Subtract 6-8i from 24-41i by subtracting corresponding real and imaginary parts.

$$Re(-\frac{1}{3}i\left(24-6+\left(-41-\left(-8\right)\right)i\right))$$

Subtract 6 from 24. Subtract -8 from -41.

$$Re(-\frac{1}{3}i\left(18-33i\right))$$

Multiply -\frac{1}{3}i times 18-33i.

$$Re(-\frac{1}{3}i\times 18-\frac{1}{3}\left(-33\right)i^{2})$$

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

$$Re(-\frac{1}{3}i\times 18-\frac{1}{3}\left(-33\right)\left(-1\right))$$

Do the multiplications in -\frac{1}{3}i\times 18-\frac{1}{3}\left(-33\right)\left(-1\right). Reorder the terms.

$$Re(-11-6i)$$

The real part of -11-6i is -11.

$$-11$$