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From \,\, where \,\, you \,\, left \,\, off: \\\\ \tan\frac{\theta}{2}\left ( \frac{\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}} \right ) = itan\frac{\theta}{2}\left (...

Congratulations to Michael and Wendy, well deserved - no doubt.

Kevin Fong ftw.

Probably, but for n > 2 - you'd assume arctan(b/a) was workable.

Thank you cyl123, clever.

I'm trying to think why, algebraically, the condition is true. There is definitely an explanation.

Also as an aside, if you tan both sides of the equation, you find yourself dealing with tan[pi/2], which is of course, undefined.

Technically x < 0 is still a true statement. For the other branch of the circle, where x > 0, the governing equation would be arg(z-i)-arg(z+i)=-pi/2 or, with some rearranging, arg(z+i)-arg(z-i)=pi/2.

The necessary condition should be x < 0.

H2SO4 + 2KOH ---> K2SO4 + 2H2O n(H2SO4) = 0.120 x 0.2 = 0.024 n(KOH) = 0.065 x 0.08 = 0.0052 Now, with 0.0052 moles KOH, there should be 0.0052/2 = 0.0026 moles H2SO4 (note mole ratio). However, we have 0.024 moles H2SO4. Therfore, the amount of H2SO4 in excess is 0.024-0.0026 =...

(\cos\theta+i\sin\theta)^{4}=\cos4\theta+i\sin4\theta \,\, [By \,\, De \,\, Moivre's \,\, Theorem] \\\\ By \,\, the \,\, Binomial \,\, Theorem, \,\, (\cos\theta+i\sin\theta)^{4}=\cos^{4}\theta+4i\sin\theta\cos^{3}\theta-6\sin^{2}\theta\cos^{2}\theta-4i\sin^{3}\theta\cos\theta+\sin^{4}\theta \\\\...

You don't need tutoring for those subjects ( necessarily ). Give it a go first and don't doubt yourself.

Maybe for MX2 this attitude will get you far enough...

LOL, I had typed something similar to this, but then thought meh. Glad someone said it xD

Ellipse/Proofs - Wikipedia, the free encyclopedia This should help.

This is because when you derive the formula of an ellipse, you end up with x^2/a^2 + y^2/a^2(1-e^2) = 1. We let b^2 = a^2(1-e^2) to make things neater and simpler, I think.

w=\frac{z-i}{z-2} \\\\ Let \,\, z=x+iy \\\\ w = \frac{x+i(y-1)}{x-2+iy} \\\\ = \frac{x+i(y-1)}{x-2+iy}\times \frac{x-2-iy}{x-2-iy} \\\\ = \frac{x(x-2)-ixy+i(x-2)(y-1)+y(y-1)}{(x-2)^{2}+y^{2}} \\\\ Re(w)=Re(\frac{x(x-2)-ixy+i(x-2)(y-1)+y(y-1)}{(x-2)^{2}+y^{2}})=0 \\\\ \therefore x(x-2)+y(y-1)=0...

z^{3}=8=8cis0 \\\\ z = 2cis\left ( \frac{2k\pi}{3} \right ), \,\, k=0,\pm 1 \\\\ z = 2, \,\, 2cis\frac{2\pi}{3}, \,\, 2cis(-\frac{2\pi}{3}) \\\\ = 2, \,\, -1+i\sqrt{3}, \,\, -1-i\sqrt{3}

Russkaya?

Where the hell did I get (a,0) from?...this is what happens when you abstain from maths for even one week.