Solve (-1+4)*(2-5i)*(2-5i)

$$( - 1 + 4 ) \cdot ( 2 - 5 i ) \cdot ( 2 - 5 i )$$

Evaluate

$$-63-60i$$

Solution Steps

Add -1 and 4 to get 3.

$$3\left(2-5i\right)\left(2-5i\right)$$

Multiply 3 times 2-5i.

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

Do the multiplications.

$$\left(6-15i\right)\left(2-5i\right)$$

Multiply complex numbers 6-15i and 2-5i like you multiply binomials.

$$6\times 2+6\times \left(-5i\right)-15i\times 2-15\left(-5\right)i^{2}$$

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

$$6\times 2+6\times \left(-5i\right)-15i\times 2-15\left(-5\right)\left(-1\right)$$

Do the multiplications.

$$12-30i-30i-75$$

Combine the real and imaginary parts.

$$12-75+\left(-30-30\right)i$$

Do the additions.

$$-63-60i$$

Real Part

$$-63$$

Solution Steps

Add -1 and 4 to get 3.

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

Multiply 3 times 2-5i.

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

Do the multiplications in 3\times 2+3\times \left(-5i\right).

$$Re(\left(6-15i\right)\left(2-5i\right))$$

Multiply complex numbers 6-15i and 2-5i like you multiply binomials.

$$Re(6\times 2+6\times \left(-5i\right)-15i\times 2-15\left(-5\right)i^{2})$$

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

$$Re(6\times 2+6\times \left(-5i\right)-15i\times 2-15\left(-5\right)\left(-1\right))$$

Do the multiplications in 6\times 2+6\times \left(-5i\right)-15i\times 2-15\left(-5\right)\left(-1\right).

$$Re(12-30i-30i-75)$$

Combine the real and imaginary parts in 12-30i-30i-75.

$$Re(12-75+\left(-30-30\right)i)$$

Do the additions in 12-75+\left(-30-30\right)i.

$$Re(-63-60i)$$

The real part of -63-60i is -63.

$$-63$$