complex numbers

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karliekay

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Sep 21, 2010
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If a+bi is a complex number but (a+bi)^3 is a real number, we can conclude

a. a=0
b. b=0
c. 3a^2b=b^3
d. a^3=3b^2

I have no clue how to start this. :?
 
Really? You can ALWAYS do it the hardest way possible...

(a+bi)^3 = something1 a^3 + something2 (a^2)bi + something3 a(bi)^2 + something4 (bi)^3

This might simplify in this way

(a+bi)^3 = something1 a^3 + something2 (a^2)bi - something3 ab^2 - something4 [(b)^3]i

Gathering all the imaginaary parts, something2 (a^2)b - something4 (b^3) = 0 <== This is the magic part. Why is this so?

This can be solved with b = 0 or something2 (a^2) - something4 (b^2) = 0

Find a simpler way to say "something" and see what you get.
 
this still doesnt help me, I have the answer key and the answer is c. I just dont know how they came up with it. I would like to know.
 
karliekay said:
If a+bi is a complex number but (a+bi)^3 is a real number, we can conclude

a. a=0
b. b=0
c. 3a^2b=b^3
d. a^3=3b^2

I have no clue how to start this. :?
Click to expand...


"expand" (a + bi)[sup:fwbtvqi6]3[/sup:fwbtvqi6] by doing this multiplication:

(a + bi)(a + bi)(a + bi)

Simplify what you get by combining like terms, and changing any of the "i[sup:fwbtvqi6]2[/sup:fwbtvqi6]" factors you get to -1.

Then, look at the fact that the result is supposed to be a REAL number, which means there should not be any factors of "i" left in the result. That means the coefficient of any "i" terms must equate to 0.

Try this approach, and SHOW us your work. It's hard to assist (without just GIVING an answer) if we can't tell where you are having trouble.
 
when I multiplied (a+bi)(a+bi) i got
a^2+2abi-b^2

is that right?

If it is right then I multiplied that by (a+bi)

I got

a^3+3a^2bi-3ab^2-b^3i

Is that right.
 
karliekay said:
when I multiplied (a+bi)(a+bi) i got
a^2+2abi-b^2

is that right?

If it is right then I multiplied that by (a+bi)

I got

a^3+3a^2bi-3ab^2-b^3i

Is that right.
Click to expand...

Yes, that's right. Now, let's group the terms containing "i" together:

a[sup:3r9kzcu7]3[/sup:3r9kzcu7] - 3ab[sup:3r9kzcu7]2[/sup:3r9kzcu7] + (3a[sup:3r9kzcu7]2[/sup:3r9kzcu7]bi - b[sup:3r9kzcu7]3[/sup:3r9kzcu7]i])

remove that common factor of "i" from the last two terms:

a[sup:3r9kzcu7]3[/sup:3r9kzcu7] - 3ab[sup:3r9kzcu7]2[/sup:3r9kzcu7] + i*(3a[sup:3r9kzcu7]2[/sup:3r9kzcu7]b - b[sup:3r9kzcu7]3[/sup:3r9kzcu7])

In order for this to be a REAL number, the coefficient of "i" must be 0....that is, 3a[sup:3r9kzcu7]2[/sup:3r9kzcu7]b - b[sup:3r9kzcu7]3[/sup:3r9kzcu7] = 0

Ok...what does that tell you about 3a[sup:3r9kzcu7]2[/sup:3r9kzcu7]b and b[sup:3r9kzcu7]3[/sup:3r9kzcu7]?
 
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