Full solution

Q. Express as a complex number in simplest a+bi form:

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\frac{16-28 i}{-4-6 i}

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Answer:

Identify complex conjugate: Identify the complex conjugate of the denominator. $\newline$ To divide complex numbers, we multiply the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of $(-4-6i)$ is $(-4+6i)$ .
Multiply by conjugate: Multiply the numerator and the denominator by the complex conjugate of the denominator. $\newline$ We have $(16-28i)/(-4-6i)$ . Multiply by $(-4+6i)/(-4+6i)$ to get rid of the imaginary part in the denominator.
Distribute and multiply numerators: Apply the distributive property to multiply out the numerators. $\newline$ Multiply $(16-28i)$ by $(-4+6i)$ . $\newline$ $(16-28i)(-4+6i) = 16(-4) + 16(6i) - 28i(-4) - 28i(6i)$
Calculate products in numerators: Calculate the products in the numerators. $\newline$ $16(-4) = -64$ $\newline$ $16(6i) = 96i$ $\newline$ $-28i(-4) = 112i$ $\newline$ $-28i(6i) = -168i^2$ (Remember that $i^2 = -1$ )
Combine like terms in numerator: Combine like terms in the numerator. $\newline$ $-64 + 96i + 112i - 168(-1)$ $\newline$ $-64 + 208i + 168$ $\newline$ $104 + 208i$
Distribute and multiply denominators: Apply the distributive property to multiply out the denominators. $\newline$ Multiply $(-4-6i)$ by $(-4+6i)$ . $\newline$ $(-4-6i)(-4+6i) = (-4)(-4) + (-4)(6i) - 6i(-4) - 6i(6i)$
Calculate products in denominators: Calculate the products in the denominators. $\newline$ $(-4)(-4) = 16$ $\newline$ $(-4)(6i) = -24i$ $\newline$ $-6i(-4) = 24i$ $\newline$ $-6i(6i) = -36i^2$ (Again, $i^2 = -1$ )
Combine like terms in denominator: Combine like terms in the denominator. $\newline$ $16 - 24i + 24i - 36(-1)$ $\newline$ $16 + 36$ $\newline$ $52$
Write in $a+bi$ form: Write the complex number in $a+bi$ form. Now we have the numerator $104 + 208i$ and the denominator $52$ . Divide both the real part and the imaginary part of the numerator by the denominator. $(104 + 208i) / 52$ $104/52 + (208i/52)$ $2 + 4i$