Melody

I have played with it but I do not know how to solve it either.    

I just answered a complex number question this morning and I suspect it was you who asked.

how about respoinding to answers that you are given !

You have to be joking.

This format is not acceptable.

\(37^2=(40-3)^2=1600-2*120+9=1369\)

How do I fing a

As I have already said.

1) I already found c in terms of a and b

2) I have already divided by (x+1)  using algebraic long division, to get the  b in terms of a

3) Now I need to divide that last answer by (x+1)^2 = x^2+2x+1   Using algebraic long division

This will give me a remiander in 'a'.  I know the remainder has to be 0 so put it equal to 0 and solve for a.

Bingo I will have my answer.

I'm not 'essentially' doing anything.

If a polynomial say f(x) is divisable by   (x+k)   then  f(-k) must be zero.     REMAINDER THEOREM

[  f(-k)=the remainder and the remainder in this case will be 0  ]

I have used this fact to help with my solution.

Are you Frosty or are you someone else?  

The complex numbers $z$ and $w$ satisfy $|z| = |w| = 1$ and $zw \ne -1.$

(a) Prove that \(\overline{z} = \frac{1}{z}\quad and \quad \overline{w} = \frac{1}{w}\)

(b) Prove that \(\frac{z + w}{zw + 1}\) is a real number.

\(Let \quad z=e^{i\theta}=cos\theta+ isin \theta\qquad\\ and \quad w=e^{i\alpha}=cos\alpha +isin \alpha\\ \theta+\alpha \ne (2n+1)\pi\\~\\ Prove \;\; \bar z=\frac{1}{z} \\\bar z=cos\theta-isin\theta\\ \frac{1}{z}=(e^{i\theta})^{-1}\\ \frac{1}{z}=e^{-\theta i}\\ \frac{1}{z}=cos(-\theta)+isin(-\theta)\\ \frac{1}{z}=cos(\theta)-isin(\theta)\\ \frac{1}{z}=\bar z \qquad QED \)

Please present you question properly.

Without unrendered LaTex

I don't understand what is being asked for.