Spesh Help...

[asdfqwerty]

hello everyone

how would i  shade the region arg(z+i) less than equal to pi/4
i know that IzI = I z-1+i I
y=x-1

thanks

[brightsky]

draw a massive open circle at (0,-1). draw in the line extending from (0,-1) which makes an angle of pi/4 with the horizontal. draw in the dotted line extending from (0,-1) horizontally to the left. shade in the 'bottom' region enclosed by the two lines drawn.

[tiff_tiff]

thank you everyone.

I was also wondering,
for cos^-1 (cos(-pi/3))
i know the domain for cos is 0 to pi, and thus the answer can't be -pi/3, so i decided to just add 2pi, but then the answer is pi/3 why?

[Sanguinne]

cos is positive in the fourth quadrant

therefore cos(-pi/3) = cos(pi/3)

that is why the answer is pi/3

[tiff_tiff]

yup i got it thank you.

I'm sorry i feel like i'm asking too much... but if anyone who is avaliable please help:

if f^-1 (x) = pi-sin^-1(x)
given that f: [-pi/2, pi/2] f(x)=2sin(x-pi) , express f^-1(x) in terms of sin^-1(x).

thank you

[tiff_tiff]

please help:
just to double check:
for the function y = sec(2x-pi/2) state the transformations to the function y = sec(x)
is it a dilation by a factor 2 from the x axis and a translation pi/4 units to the right?

[Zealous]

please help:
just to double check:
for the function y = sec(2x-pi/2) state the transformations to the function y = sec(x)
is it a dilation by a factor 2 from the x axis and a translation pi/4 units to the right?

Hi.

It's a dilation by factor 1/2 away from the y axis then a translation of pi/4 units to the right. You just got dilation the wrong way round.

[tiff_tiff]

thank you !!!

this is sorta a big turn but i also have a vector question:
OABC is a parallelogram. D is a point on CB such that CD:DB = p:q (D is closer to C than B)
OD intersects AC at R
OA = a
OC = c
Given that OR = k1OD and CR = k2CA and using OC + RC = 0R find k1 and k2 in terms of p and q

urgent please and thank you so much

[brightsky]

OD = OC + CD = OC + p/(p+q) CB = c + p/(p+q) a = c + p/(p+q) a

Hence:

OR = k1 OD = k1 c + k1p/(p+q) a

But:

OR = OC + CR = OC + k2 CA = c + k2 (a - c) = (1 - k2) c + k2 a

Equating coefficients:

k1 = 1 - k2...(1)
k1 p/(p+q) = k2...(2)

Sub (2) into (2):

k1 = 1 - k1 p/(p+q)
k1 [1 + p/(p+q)] = 1
k1 = 1/[1 + p/(p+q)] = (p+q)/(2p+q)

Sub this back into (2):

k2 = [(p+q)/(2p+q)] [p/(p+q)] = p/(2p+q)

Hence, k1 = (p+q)/(2p+q) and k2 = p/(2p+q).