Unit 58: Imaginary Numbers

Imaginary numbers on the SAT are represented by the letter i .  They will almost always tell you that \(i=\sqrt{-1}\) .  Because you cannot take the square root of a negative number, it is imaginary.

There are two types of imaginary number problems on the SAT.

  1. Combining like terms
  2. in the denominator of a fraction

The first kind is rather easy.  You are given a combination of real and imaginary numbers and asked to combine them.  In this case, you treat i just like a variable.

An example:

Which of the following is equivalent to \(\left(7-3i\right)-(4-5i)\)?

A) \(3-8i\)
B) \(3+2i\)
C) \(11-5i\)
D) \(11+2i^2\)

Combining our like terms gives us

$$3+2i$$

That’s it!

A small wrinkle may be where they give you an \(I^2\).  Because \(i=\sqrt{-1}\), \(i^2=\sqrt{-1}\sqrt{-1}\) or just -1.  In general you can always just substitute -1 for \(\boldsymbol i^{\mathbf2}\).

$$\left(4-3i^2\right)-i(4-i)$$

Would convert to 

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

On any i problem, always be very, very careful with signs!  They love to give you answers that look correct if you flipped a sign along the way.

The second kind of imaginary number problem, when you have i in the denominator, is a little more complicated. 

An example will make this clearer.

If \(i=\sqrt{-1}\), then \(\frac{3+6i}{2-4i}\)  is equal to which of the following?

A) \(\frac{3i}2-\frac64\)
B) \(\frac{2i}5-\frac9{10}\)
C) \(\frac{6i}5-\frac9{10}\)
D) \(\frac{10i-9}{10}\)

Strategy:

  1. Multiply the denominator and the numerator by the flipped sign version of the denominator.
  2. Foil.
  3. Replace all \(\boldsymbol i^{\mathbf2}\)s with \(\boldsymbol(\boldsymbol-\mathbf1\boldsymbol)\)

We have our denominator with i in it.  The first thing we should do is multiply the top and bottom by the denominator with its sign flipped.

$$\frac{3+6i}{2-4i}\;\times\;\frac{2+4i}{2+4i}$$

Because we are multiplying by a number over itself, it is the same as multiplying by 1, which doesn’t mathematically change the equation we are working with.  However, by doing this, when we FOIL the denominator, we will ALWAYS be able to cancel out the middle i terms, and get an \(I^2\), which we can then convert to -1 and make a real number.  

We’ll multiply through the denominators first.

$$\left(2-4i\right)\left(2+4i\right)=4+8i-8i-16i^2\\4-16i^2=4-16(-1))=4+16=20$$

We multiply through the numerators in the same fashion.

$$\left(3+6i\right)\left(2+4i\right)=\;6+12i+12i+24i^2\\6+24i-24\\24i-18$$

Putting it all together, we would have:

$$\frac{24i-18}{20}=\frac{24i}{20}-\frac{18}{20}=\frac{6i}5-\frac9{10}$$

That’s as far as we can reduce it down.