Complex Numbers Pre- Calculus

[tayhalo]

Write the expression in the form a+bi , where a and b are real numbers.

((9/2)-(?3/2)i)^3

 

[mmm4444bot]

((9/2)-(?3/2)i)^3

Your typing means this:

\(\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right)^{3}\)

Is this what you intend ?

Can you determine the following ?

i^3 =

(sqrt[3]/2)^3 =

 

[tayhalo]

No the parentheses are around the entire problem

 

[mmm4444bot]

Please excuse me. I misread the grouping symbols. (We do not need parentheses around 9/2)

(9/2 - sqrt(3)/2 i)^3

So, let's start with the following.

(9/2 - sqrt(3)/2 i)^2

What do you get, when you apply the FOIL algorithm to this square ?

 

[tayhalo]

81/4-9sqrt(3)/4-9sqrt(3)/4+3/4

 

[mmm4444bot]

81/4 - 9sqrt(3)/4 - 9sqrt(3)/4 + 3/4

What happened to the factors of i ?

The rest of it is okay.

\(\displaystyle \left ( \frac{9}{2} \right ) \cdot \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \;=\; - \frac{9 \ \sqrt{3}}{4} \cdot i\)

\(\displaystyle \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \cdot \left ( - \frac{\sqrt{3}}{2} \cdot i \right ) \;=\; \frac{3}{4} \cdot i^2\)

Do you know the value of i^2

Your result above can be simplified, after you replace i^2 with its value.

 

[tayhalo]

I have no idea, and the book has no example problem.

 

[mmm4444bot]

Your math text must define the symbol i somewhere. I think you just missed it.

i = sqrt(-1)

Therefore, i^2 = -1

In other words, replace i^2 with -1 in your foiled result above, and simplify.

(Also, don't forget to fix the other missing factor of i that I showed you.)

Please post your corrections, and we'll go from there.

 

[tayhalo]

78/4 - 9sqrt(3)/4i - 9sqrt(3)/4i

 

[mmm4444bot]

Okay, you're getting closer.

The Rational number 78/4 can be reduced, yes ?

Also, the two terms -9 sqrt(3) i/4 and -9 sqrt(3) i/4 are like-terms.

I mean, they're identical, yes ?

Combine them.

With these two simplications, we end up with the following square.

\(\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right )^{2} \;=\; \frac{39}{2} \;-\; \frac{9 \ \sqrt{3}}{2} \cdot i\)

The original expression has three factors (a cube), and you've now multiplied two of them together (a square).

Therefore, to finish, multiply the squared result by the third factor and you're done.

\(\displaystyle \left ( \frac{9}{2} \;-\; \frac{\sqrt{3}}{2} \cdot i \right ) \left( \frac{39}{2} \;-\; \frac{9 \ \sqrt{3}}{2} \cdot i \right )\)