Solve (-3+5texttt{i})-(4-2texttt{i})(1+6texttt{i})

$$(-3+5 \texttt{i} )-(4-2 \texttt{i} )(1+6 \texttt{i} )$$

Evaluate

$$-19-17i$$

Solution Steps

Multiply complex numbers 4-2i and 1+6i like you multiply binomials.

$$-3+5i-\left(4\times 1+4\times \left(6i\right)-2i-2\times 6i^{2}\right)$$

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

$$-3+5i-\left(4\times 1+4\times \left(6i\right)-2i-2\times 6\left(-1\right)\right)$$

Do the multiplications in 4\times 1+4\times \left(6i\right)-2i-2\times 6\left(-1\right).

$$-3+5i-\left(4+24i-2i+12\right)$$

Combine the real and imaginary parts in 4+24i-2i+12.

$$-3+5i-\left(4+12+\left(24-2\right)i\right)$$

Do the additions in 4+12+\left(24-2\right)i.

$$-3+5i-\left(16+22i\right)$$

Subtract 16+22i from -3+5i by subtracting corresponding real and imaginary parts.

$$-3-16+\left(5-22\right)i$$

Subtract 16 from -3. Subtract 22 from 5.

$$-19-17i$$

Real Part

$$-19$$

Solution Steps

Multiply complex numbers 4-2i and 1+6i like you multiply binomials.

$$Re(-3+5i-\left(4\times 1+4\times \left(6i\right)-2i-2\times 6i^{2}\right))$$

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

$$Re(-3+5i-\left(4\times 1+4\times \left(6i\right)-2i-2\times 6\left(-1\right)\right))$$

Do the multiplications in 4\times 1+4\times \left(6i\right)-2i-2\times 6\left(-1\right).

$$Re(-3+5i-\left(4+24i-2i+12\right))$$

Combine the real and imaginary parts in 4+24i-2i+12.

$$Re(-3+5i-\left(4+12+\left(24-2\right)i\right))$$

Do the additions in 4+12+\left(24-2\right)i.

$$Re(-3+5i-\left(16+22i\right))$$

Subtract 16+22i from -3+5i by subtracting corresponding real and imaginary parts.

$$Re(-3-16+\left(5-22\right)i)$$

Subtract 16 from -3. Subtract 22 from 5.

$$Re(-19-17i)$$

The real part of -19-17i is -19.

$$-19$$