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Complex Numbers Question (1 Viewer)
- Thread starter midifile
- Start date
ok its late, but a quick glance, and i picked up the question which is rather generic and which you realy should learn how to do. 
question 13) its wrong, so maybe that's why you had problems...it should read show that z^n + 1/z^n=2cos n theta, and from that, its pretty generic. (hint, demoevres)
with 17 part c), im assuming the roots are tan pi/16, tan 3pi/16, tan 5pi/16 and tan 7pi/16.
from there....you know that a^2 + b^2 = (a+b)^2 - 2ab...
that works for a^2+b^2+c^2 to however many you want...
so basically, you want
the sum of the roots squared MINUS two times the product by pairs.
in other words, (-4)^2 - (2)(-6) = 28.
with 20 a), you know that tan alpha =1/3, so tan 2 alpha=3/4 (using the tan 2alpha expansion), and then you do it again and you get tan 4alpha =24/7; then you take the inverse tan of both sides and you're there.
b) and c) i'll look over later.
24 i)...that seems rather basic. mark the point x=-2, then draw a circle around it with radius 1. then from there, its pretty obvious that the longest distance from the origin is 3 and the shortest is 1. you can prove that with trig.
get back on part ii), that requires my brain to be working. xD
soz, my brain's too fried to look at anything more complex that the ones i did. hope it helps.
question 13) its wrong, so maybe that's why you had problems...it should read show that z^n + 1/z^n=2cos n theta, and from that, its pretty generic. (hint, demoevres)
with 17 part c), im assuming the roots are tan pi/16, tan 3pi/16, tan 5pi/16 and tan 7pi/16.
from there....you know that a^2 + b^2 = (a+b)^2 - 2ab...
that works for a^2+b^2+c^2 to however many you want...
so basically, you want
the sum of the roots squared MINUS two times the product by pairs.
in other words, (-4)^2 - (2)(-6) = 28.
with 20 a), you know that tan alpha =1/3, so tan 2 alpha=3/4 (using the tan 2alpha expansion), and then you do it again and you get tan 4alpha =24/7; then you take the inverse tan of both sides and you're there.
b) and c) i'll look over later.
24 i)...that seems rather basic. mark the point x=-2, then draw a circle around it with radius 1. then from there, its pretty obvious that the longest distance from the origin is 3 and the shortest is 1. you can prove that with trig.
get back on part ii), that requires my brain to be working. xD
soz, my brain's too fried to look at anything more complex that the ones i did. hope it helps.
Thanks so muchu-borat said:ok its late, but a quick glance, and i picked up the question which is rather generic and which you realy should learn how to do.
question 13) its wrong, so maybe that's why you had problems...it should read show that z^n + 1/z^n=2cos n theta, and from that, its pretty generic. (hint, demoevres)
with 17 part c), im assuming the roots are tan pi/16, tan 3pi/16, tan 5pi/16 and tan 7pi/16.
from there....you know that a^2 + b^2 = (a+b)^2 - 2ab...
that works for a^2+b^2+c^2 to however many you want...
so basically, you want
the sum of the roots squared MINUS two times the product by pairs.
in other words, (-4)^2 - (2)(-6) = 28.
with 20 a), you know that tan alpha =1/3, so tan 2 alpha=3/4 (using the tan 2alpha expansion), and then you do it again and you get tan 4alpha =24/7; then you take the inverse tan of both sides and you're there.
b) and c) i'll look over later.
24 i)...that seems rather basic. mark the point x=-2, then draw a circle around it with radius 1. then from there, its pretty obvious that the longest distance from the origin is 3 and the shortest is 1. you can prove that with trig.
get back on part ii), that requires my brain to be working. xD
soz, my brain's too fried to look at anything more complex that the ones i did. hope it helps.
anything is better than nothing