Posts tagged with: complex numbers

This may be a little advanced for the SAT, but complex numbers sometimes show up –as do cubic polynomials– so hopefully you can address this for me! TIA!

Which of the following could be the full set of complex roots of a cubic polynomial with real coefficients?

A. { 0, 1, i}
B. {1, i, 2i}
C. {2, i}
D. {3, 2 + i, 2 – i}

 

I don’t feel like I’m qualified to teach a lesson on this. I know that a cubic polynomial must have 1 or 3 real roots, but I’m too far removed from the class I learned that in to reconstruct the proof here. (A lesson that’s always worth relearning in SAT prep: not all high school math is SAT math.) However, I can still help with this question because it’s multiple choice and happens to lend itself to one of my favorite approaches: we can backsolve!

If you translate the answer choices to what the factors of the polynomial would be, you can multiply those and see which one ends up with all real coefficients. I gravitated right to choice D, because that 2 + i and 2 – i conjugate looks like it’ll cancel out nicely. Let’s see!

We know that a cubic polynomial with roots 3, 2 + i, 2 – i would have the factored form: y=(x-3)(x-(2+i))(x-(2-i)). Now FOIL the complex factors:

y=(x-3)(x-(2+i))(x-(2-i))
y=(x-3)(x^2-(2+i)x-(2-i)x+(2+i)(2-i))
y=(x-3)(x^2-2x-ix-2x+ix+(4-i^2)) <-- All the i terms will go away here!
y=(x-3)(x^2-4x+(4-(-1)))
y=(x-3)(x^2-4x+(4+1))
y=(x-3)(x^2-4x+5)

See how the i terms cancel out? That won’t happen in any other choice, which means the other choices will result in coefficients that aren’t real. We don’t even need to fully expand this one to know it’s the right move.

If you remember that the square of i is –1, you can reduce these to purely real numbers before you even FOIL.

image

In a + bi form, 5 is just 5 + 0i, so a + b will be 5 + 0 = 5.

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Test 4 Section 3 #14

When you’re dividing complex numbers, you have to multiply the top and bottom of the fraction by the complex conjugate of the bottom. This creates a real number in the bottom of the fraction, which is awesome.

The bottom of this fraction is 3-2i, so its complex conjugate is 3+2i. To get started, then, we write the following:

    \begin{align*}\dfrac{8-i}{3-2i}\times\dfrac{3+2i}{3+2i}\end{align*}

Then we simplify as much as we can. First we FOIL:

    \begin{align*}&\dfrac{(8-i)(3+2i)}{(3-2i)(3+2i)}\\=&\dfrac{24+16i-3i-2i^2}{9+6i-6i-4i^2}\\=&\dfrac{24+13i-2i^2}{9-4i^2}\end{align*}

Now, remembering that i=\sqrt{-1} and therfore that i^2=-1, we make that substitution and simplify further:

    \begin{align*}&\dfrac{24+13i-2(-1)}{9-4(-1)}\\=&\dfrac{24+13i+2}{9+4}\\=&\dfrac{26+13i}{13}\\=&2+i\end{align*}

So there you have it. Once we simplify that complex fraction into a+bi form, we see that it’s just 2+i, which means a=2.