Complex Question and another question (1 Viewer)

[DraconisV]

Ok i have this complex number question that im confused with.

Ok.

1. Solve the equation Z^5 + 81Z = 0, expressing the answers in the form a + ib, where a and b are real.

Ok im confused can someone show me how this is done.

Also

2. The graph y = (ax^2 + bx + c)/(x^2 + qx + r) has line x=1, x=3 and y=2 as the asymptotes and (0,1) as turning point.

Use this information to show that y = (2x^2 - 4x +3)/(x^2 - 4x + 3)

Im totally lost with this one, ive tried doing all sorts of things but nothing has made me progress at all.

Can someone please show me, very closely how the hell this dam bloody question is done.

Thank you guys.

 

[pLuvia]

First one, factorise it, then use DeMoivre's theorem to get the four roots of unity. Then you should end up with four equations then just change them back to a+bi form

Second one, you should know that the denominator when equalled to zero are yields the horizontal asymptotes i.e. y=2 and that the numerator yields the vertical asymptotes i.e. x=1, x=3. So you can sub those numbers into those equations, and then using the first derivative and subbing in the (0,1) you have a second equation, then solve simultaneously for the pronumerals

Hope that helped

 

[Trev]

Since it's 'x' assymptotes are x=1 and x=3 that means the denominator (as it can't equal zero) will be (x-1)(x-3)=x²-4x+3.

You are given (0,1) as a point, so sub this into the equation you have so far:
1=c/r, and you figured out r=3, so c=3.
You now have y=(ax² - bx +3)/(x^2 - 4x + 3)
Differentiate this to get:
dy/dx=[(x²-4x+3)(2ax+b)-(ax²+bx+3)(2x-4)]/[(x²-4x+3)²]
Since (0,1) is a turning point, then when dy/dx=0, x=0. So:
0=3b+12, b=-4.

(I can't be bothered typing the rest) but, make 'x' the subject and do the same as for the 'x' assymptoes, the denominator should equal zero when you sub in y=2, so you will then find out the value for a.