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From Coroneos, Let \; f(\theta )=cos\theta + isin \theta \\ \therefore f'(\theta)=-sin\theta+icos\theta=i(cos\theta+isin\theta)=if(\theta) \\ i.e \; \frac{f'(\theta)}{f(\theta)}=i, and \therefore \int\frac{f'(\theta)}{f(\theta)}= \int i d \theta \\ ln(f(\theta))= i\theta + C \\ When...

1/(3x+1)>1/(2x-3) (2x-3)^2(3x+1)>(3x+1)^2(2x-3) (2x-3)^2(3x+1)-(3x+1)^2(2x-3)>0 (2x-3)(3x+1)(2x-3-3x-1)>0 -(2x+3)(3x+1)(x+4)>0 by graphing or testing value we get x<-4, -1/3 < x < 3/2

Solving graphically yields x<1, x>2. There is also the case method. for (2x-3)/(x-2)>1 either x<2 or x>2 case 1; when x<2, (2x-3)/(x-2)>1 2x-3 < x-2 x<1 case 2; when x>2 (2x-3)/(x-2)>1 2x-3>x-2 x>1, but x>2 therefore solution to this inequality is x>2 it follows...

I dont think inverse functions is quite the reason why this works. if this was the case then if i wanted to find a polynomial with roots 1/(alpha^2), etc.. then i woould sub in sqrt(1/alpha). which leads to a polynomial fo degree four and hence is obviously wrong. (assuming my algebra wasn't...

di) This is simply the locus of a circle with radius three and centre (3,0) in the argand diagram. ii) \textrm{let A represent the point (3,0) or the complex number 3+0i.}\\ \textrm{ Now, OA=AP (radii of circle)} \\ \therefore \triangle OAP \textrm{is isosceles, and} \angle AOP= \angle APO\\...