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# P4 Questions watch

1. Could someone show me the proof for the following question. I tried it and couldn't do it and then when I tried again it seemed to go on forever! Thanks:

1) Given that T = (x-yi)/(x+yi), where x, y are members of R, show that:

(1 + T^2)/2T = (x^2 - y^2)/(x^2 + y^2)

Secondly, could someone confirm the module and argument of the following: z=(2+i)(3-2i). The answer is supposed to be: |z|= sqrt 65 and arg z = -0.421
however, I can't get the same answer! Thanks
2. 1)

T=(x-yi)/(x+yi)

T^2 = (x^2 - y^2 -2xyi)/(x^2 - y^2 +2xyi)

T^2 + 1 = (x^2 - y^2 -2xyi)/(x^2 - y^2 +2xyi) + (x^2 - y^2 +2xyi)/(x^2 - y^2 +2xyi)

T^2 + 1 = (2x^2 - 2y^2)/(x^2 - y^2 +2xyi)

T^2 + 1 = (2x^2 - 2y^2)/(x+yi)^2

(T^2 + 1)/2T = (x^2 - y^2)/(x+yi)^2 * (x+yi)/(x-yi)

(T^2 + 1)/2T = (x^2 - y^2)/(x+yi) * 1/(x-yi)

(T^2 + 1)/2T = (x^2 - y^2)/(x^2 + y^2)
3. I get the modulus of that number to be the square root of 5/13 and the argument to be arctan[7/4].
4. (Original post by AntiMagicMan)
I get the modulus of that number to be the square root of 5/13 and the argument to be arctan[7/4].
arghh! that wasn't what I got either! so now im confused! can some more ppl reply!
5. (Original post by Hoofbeat)
arghh! that wasn't what I got either! so now im confused! can some more ppl reply!
z = (2+i)(3-2i)
z = 6 - 4i +3i -2i²
z = 8 - i
|z| =√(64+1)
|z| = 65
======

Arg(z) = arctan(-1/8)
==================
6. (Original post by Hoofbeat)
Secondly, could someone confirm the module and argument of the following: z=(2+i)(3-2i). The answer is supposed to be: |z|= sqrt 65 and arg z = -0.421
however, I can't get the same answer! Thanks
z = (2+i)(3-2i)
z = (2)(3) + (2)(-2i) + (i)(3) + (i)(-2i)
z = 6 - 4i + 3i -2i^2 = 6 - 4i + 3i + (-2)(-1)
z = 6 - i + 2
z = 8 - i

|z| = sqrt(8^2 + (-1)^2)
|z| = sqrt(64 + 1) = sqrt 65

on the argand diagram, z is in the bottom right quadrant, so the principle argument is:

tan (arg z) = -1/8
(arg z) = arctan (-1/8)
arg z = -0.124 radi8ans (3 s.f.)

first one looks weird,
7. (Original post by AntiMagicMan)
1)

T=(x-yi)/(x+yi)

T^2 = (x^2 - y^2 -2xyi)/(x^2 - y^2 +2xyi)

T^2 + 1 = (x^2 - y^2 -2xyi)/(x^2 - y^2 +2xyi) + (x^2 - y^2 +2xyi)/(x^2 - y^2 +2xyi)

T^2 + 1 = (2x^2 - 2y^2)/(x^2 - y^2 +2xyi)

T^2 + 1 = (2x^2 - 2y^2)/(x+yi)^2

(T^2 + 1)/2T = (x^2 - y^2)/(x+yi)^2 * (x+yi)/(x-yi)

(T^2 + 1)/2T = (x^2 - y^2)/(x+yi) * 1/(x-yi)

(T^2 + 1)/2T = (x^2 - y^2)/(x^2 + y^2)
Thanks that's a lot easier than how I tried to prove it! I ended up undoing all the brackest and getting confused.
8. Cheers Mik1a and Fermat, don't know where I went wrong as I used the same method as you. Obviously I can't do simple multipication or addition!

Thanks for youre help to Antimagicman!
9. (Original post by Hoofbeat)
arghh! that wasn't what I got either! so now im confused! can some more ppl reply!
Oh I'm sorry I read (2+i)(3-2i) as (2+i)/(3-2i) because I was still thinking about (x-yi)/(x+yi).

Now I do get |z| = root(65) and arg(z) = -arctan(1/8), but hey at least my first answer was right if z = (2+i)/(3-2i) .
10. (Original post by AntiMagicMan)
Oh I'm sorry I read (2+i)(3-2i) as (2+i)/(3-2i) because I was still thinking about (x-yi)/(x+yi).

Now I do get |z| = root(65) and arg(z) = -arctan(1/8), but hey at least my first answer was right if z = (2+i)/(3-2i) .
That's ok! Thanks neways! You're proof was greta, i hadn't thought of substituting 1 for (x^2 - y^2 +2xyi)/(x^2 - y^2 +2xyi)
11. (Original post by Hoofbeat)
That's ok! Thanks neways! You're proof was greta, i hadn't thought of substituting 1 for (x^2 - y^2 +2xyi)/(x^2 - y^2 +2xyi)
Ah yeah, that is a very handy thing to do if you ever need to add 1 or any integer to a quotient.

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