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Now we're into the territory of what qualifies as necessity.

need ( P ) Pronunciation Key (nd) n. A condition or situation in which something is required or wanted: crops in need of water; a need for affection. Something required or wanted; a requisite: “Those of us who led the charge for these women's issues... shared a common vision in the...

That's what lower case pi comes out as for me :(. π pi, Π Pi (It looks like a small capital)

You could use the ASTC quadrant thing for this I geuss but essentially: cos( 3π/2 + x) = cos[2π -(π/2 -x)] =cos(π/2-x) since cos is positive in the fourth quadrant and cos(x) = cos(2π-x) =sinx

You can use the [All Stations To Central] quadrants thing or you could just imagine a sin graph (I assume you're talking about the second solution where 0 < x < 360&deg; ). Sin(90-x)=cosx but I assume you meant sin(180-x)=sinx which does work. So you get x = 60&deg;, 120&deg; otherwise...

What if someone needs 99.xx?

You don't need to use absolute values, it just seems a quicker way to generate both solutions and I geuss it seems 'proper' when you're dealing with distances. Otherwise you could just use (x-2) and (2-x) as values for the distance between (x,y) and the line x=2 [same deal as using absolute...

I'm pretty sure that the distance of a point (x,y) from the line y=-5 is |y+5| and that the distance of (x,y) from x=2 is just |x-2| so |y+5| = 3/4|x-2| -----> 3x -4y -26 = 0 or 3x +4y +14=0 (EDIT: As stated above :p)

That's pretty solid, nice work .

3) According to planck the energy packed in each quantum (photon) is E = hf where h is planck's constant and f is the frequency of the photon. This means that the higher the frequency, the greater the energy i.e UV light has more energy than red light and gamma rays have more energy than x-rays...

There's probably a way but I havn't seen a conclusive algebraic proof for these types of question. You just need logic and a bit of geometry. You observe that the case of arg(z-i) = arg(z+i) + &pi;/2 holds true to the left of the im-axis and that since the difference in args is 90&deg...

I don't :p , otherwise I just look at the textbook.

Call me wrong, but isn't that just trying to show that n &ge; 1 (after subtracting 2n from both sides) given the condition that n is an integer where n is &ge; 1 ? It seems as though it's proven from the outset.

By using the sums as products thing. Consider the following: (1) cos(&theta; + &phi; ) = cos&theta;cos&phi; - sin&theta;sin&phi; (2) cos(&theta; - &phi; ) = cos&theta;cos&phi; + sin&theta;sin&phi; by adding (1) and (2) cos (&theta; + &phi; ) + cos (&theta; - &phi; ) =...

2cos5xcos4x = cosx + cos9x &int;cos5xcos4xdx = 1/2 &int; cosx + cos9x dx = 1/2sinx +1/18.sin9x +c

True . . . . . . Arg(z₁z₂) = argz₁ + arg z₂ Arg (z₁/z₂) = argz₁ - arg z₂ or so I have been led to believe. since if z₁= r₁cis&theta;₁ and z₂ =...

I have to say, I expected something different when I opened this thread. Where's the Motown at??

In which case I'm thinking it would just be vector Q - vector P. P = z = x + iy Q = -iz = y - ix PQ = Q(y - ix) - P(x +iy) = y - x - i(y + x) I don't know if that's the way to go about it, I'm starting to feel rather rusty with the complex stuff (quite prematurely, might I add).

You don't know how proud I am of that :D. Spread the word!!! P.S. For ii) was there anything to indicate that PQ was perpindicular to the x axis. Also, how did they justify gving you no marks? What did they want?

i) If P represents z then Q is a 90&deg; rotation of z clockwise which is equivalent to multiplying z by '-i'. Hence Q can be expressed as: '-iz'