Tutors Answer Your Questions about Complex Numbers (FREE)
Question 518014: Please help me with this problem:
(4i)/(3+i)
So far, I've got:
[(4i)/(3+i)] * (3-i) = (12i-4i^2)/(9-i^2)
I know that having i^2 = -1 but I was wondering if since it's already -4i^2 does that make it +4 or does it stay -4?
Thanks for your help. It is much appreciated.
Click here to see answer by MathLover1(6634)  |
Question 518014: Please help me with this problem:
(4i)/(3+i)
So far, I've got:
[(4i)/(3+i)] * (3-i) = (12i-4i^2)/(9-i^2)
I know that having i^2 = -1 but I was wondering if since it's already -4i^2 does that make it +4 or does it stay -4?
Thanks for your help. It is much appreciated.
Click here to see answer by Alan3354(30993)  |
Question 521802: I'm working on Adv. Alegbra II homework. I've done all of the problums exept for "Solving each Equation" questions; There are three equations that i don't understand how to solve. They are the following:
(#1) -6x^2-30=0
(#2) 3x^2+18=0
(#3) 2/3x^2+30=0
If you could help ASAP that would be awesome. :)
Click here to see answer by umnikmo(75) |
Question 526662: please help me urgently....
1)Find m and n if (x-2) and (x+3) are both factors of x^3+mx^2-nx-6
2) When p(x)=x^4-3x^3+ax^2+bx-6 is divided by (x+1)it has a remainder of 8. If (x-3) is a factor of p(x) find a and b.
3) If x^2-4 is a factor of 2x^3-ax^2+bx+4 find a and b.
Click here to see answer by KMST(1869)  |
Question 529655: Hello, at first we have the complex number z = -1 + i sqrt(3).
Secondly we have the fraction z/(1 + 2i).
If we substitute z in the numerator we get: {(-1 + i sqrt(3)}/(1+2i).
How do we calculate this fraction into the form a + bi, (a,b belong to R)?
Thanks a lot in advance
Click here to see answer by scott8148(6628)  |
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