Write the expression in standard form 3/3-12i I don't understand how to find out the answer
As you have written it: 1 - 12i If you mean \(\dfrac{3}{3+12i}\), you have not written that and you must use more parentheses to say what you mean.
no its 3-12i
Pardon my typo, but I still don't know which expression you intended. You WROTE 1 - 12i, since 3/3 = 1. Is that what you intended?
honestly I don't know how to do this type of problem, it's the first time I've seen it
Have you studied the Order of Operations?
its been a while
Small refresher, then. 3/3-12i = \(\dfrac{3}{3} - 12i = 1 - 12i\) 3/(3-12i) = \(\dfrac{3}{3-12i}\) Do you see the difference and why the parentheses are important?
Yes
I'm guessing that you mean the latter problem, since it is quite a bit more interesting. Standard form is a + bi, no giant fractions. Our task is to get rid of the big fraction. Normally, this is by applying the "Complex Conjugate" of the denominator. Can you write the complex conjugate of 3 - 12i?
would it be 3+12i?
Perfect! However, I don't want to use that. I see something else in this problem that will make the world a better place. EVERYTHING in there has a factor of 3. We can simplify the problem if we get rid of it. \(\dfrac{3}{3 - 12i} = \dfrac{3}{3(1 - 4i)} = \dfrac{3}{3}\cdot\dfrac{1}{1-4i} = \dfrac{1}{1-4i}\) Do you see all that?
I do
ALWAYS be in the lookout for that sort of thing. Okay, now we need the Complex Conjugate of 1 - 4i. You know already that it is 1 + 4i. Use it like this. \(\dfrac{1}{1-4i}\cdot\dfrac{1+4i}{1+4i} = \dfrac{1+4i}{(1-4i)(1+4i)}\) Multiply those two factors in the denominator and see if anything helpful occurs.
I ended up with 1-16i^2, would I chang -16i^2 to positive 17 or -17?
\(1 - 16i^2 = 1 - 16(-1) = 1 + 16 = 17\) Now what? We're almost done.
\[\frac{ 1 }{ 17 }-\frac{ 4 }{ 17 }i\] ?
If you had written + rather than -, you would be done! Hienoa!!
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