Solve 2*(5-2i)+[3*(4*3i)-2*(5-2i)]*(-2-i)

$$2 \cdot ( 5 - 2 i ) + [ 3 \cdot ( 4 \cdot 3 i ) - 2 \cdot ( 5 - 2 i ) ] \cdot ( - 2 - i )$$

Evaluate

$$70-74i$$

Solution Steps

Multiply 2 times 5-2i.

$$2\times 5+2\times \left(-2i\right)+\left(3\times 4\times \left(3i\right)-2\left(5-2i\right)\right)\left(-2-i\right)$$

Do the multiplications in 2\times 5+2\times \left(-2i\right).

$$10-4i+\left(3\times 4\times \left(3i\right)-2\left(5-2i\right)\right)\left(-2-i\right)$$

Multiply 3 and 4 to get 12.

$$10-4i+\left(12\times \left(3i\right)-2\left(5-2i\right)\right)\left(-2-i\right)$$

Multiply 12 and 3i to get 36i.

$$10-4i+\left(36i-2\left(5-2i\right)\right)\left(-2-i\right)$$

Multiply 2 times 5-2i.

$$10-4i+\left(36i-\left(2\times 5+2\times \left(-2i\right)\right)\right)\left(-2-i\right)$$

Do the multiplications in 2\times 5+2\times \left(-2i\right).

$$10-4i+\left(36i-\left(10-4i\right)\right)\left(-2-i\right)$$

Subtract 10-4i from 36i by subtracting corresponding real and imaginary parts.

$$10-4i+\left(10+\left(36-\left(-4\right)\right)i\right)\left(-2-i\right)$$

Subtract -4 from 36.

$$10-4i+\left(-10+40i\right)\left(-2-i\right)$$

Multiply complex numbers -10+40i and -2-i like you multiply binomials.

$$10-4i-10\left(-2\right)-10\left(-i\right)+40i\left(-2\right)+40\left(-1\right)i^{2}$$

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

$$10-4i-10\left(-2\right)-10\left(-i\right)+40i\left(-2\right)+40\left(-1\right)\left(-1\right)$$

Do the multiplications in -10\left(-2\right)-10\left(-i\right)+40i\left(-2\right)+40\left(-1\right)\left(-1\right).

$$10-4i+20+10i-80i+40$$

Combine the real and imaginary parts in 20+10i-80i+40.

$$10-4i+20+40+\left(10-80\right)i$$

Do the additions in 20+40+\left(10-80\right)i.

$$10-4i+\left(60-70i\right)$$

Combine the real and imaginary parts.

$$10+60+\left(-4-70\right)i$$

Do the additions.

$$70-74i$$

Real Part

$$70$$

Solution Steps

Multiply 2 times 5-2i.

$$Re(2\times 5+2\times \left(-2i\right)+\left(3\times 4\times \left(3i\right)-2\left(5-2i\right)\right)\left(-2-i\right))$$

Do the multiplications in 2\times 5+2\times \left(-2i\right).

$$Re(10-4i+\left(3\times 4\times \left(3i\right)-2\left(5-2i\right)\right)\left(-2-i\right))$$

Multiply 3 and 4 to get 12.

$$Re(10-4i+\left(12\times \left(3i\right)-2\left(5-2i\right)\right)\left(-2-i\right))$$

Multiply 12 and 3i to get 36i.

$$Re(10-4i+\left(36i-2\left(5-2i\right)\right)\left(-2-i\right))$$

Multiply 2 times 5-2i.

$$Re(10-4i+\left(36i-\left(2\times 5+2\times \left(-2i\right)\right)\right)\left(-2-i\right))$$

Do the multiplications in 2\times 5+2\times \left(-2i\right).

$$Re(10-4i+\left(36i-\left(10-4i\right)\right)\left(-2-i\right))$$

Subtract 10-4i from 36i by subtracting corresponding real and imaginary parts.

$$Re(10-4i+\left(10+\left(36-\left(-4\right)\right)i\right)\left(-2-i\right))$$

Subtract -4 from 36.

$$Re(10-4i+\left(-10+40i\right)\left(-2-i\right))$$

Multiply complex numbers -10+40i and -2-i like you multiply binomials.

$$Re(10-4i-10\left(-2\right)-10\left(-i\right)+40i\left(-2\right)+40\left(-1\right)i^{2})$$

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

$$Re(10-4i-10\left(-2\right)-10\left(-i\right)+40i\left(-2\right)+40\left(-1\right)\left(-1\right))$$

Do the multiplications in -10\left(-2\right)-10\left(-i\right)+40i\left(-2\right)+40\left(-1\right)\left(-1\right).

$$Re(10-4i+20+10i-80i+40)$$

Combine the real and imaginary parts in 20+10i-80i+40.

$$Re(10-4i+20+40+\left(10-80\right)i)$$

Do the additions in 20+40+\left(10-80\right)i.

$$Re(10-4i+\left(60-70i\right))$$

Combine the real and imaginary parts in 10-4i+60-70i.

$$Re(10+60+\left(-4-70\right)i)$$

Do the additions in 10+60+\left(-4-70\right)i.

$$Re(70-74i)$$

The real part of 70-74i is 70.

$$70$$