hard but not impossible complex q. (1 Viewer)

[YannY]

gm.haider - the answer is a curve hence it is a locus. if i give you the modulus version of the locus then it would be the same as giving you the answer, besides i'll need to make up the modulus which takes effort and time so no can do mate.

No it is not z=i lemme sub that in for you.

Arg(z(z+3)) when z=i
=Arg(i(i+3))
=Arg i + arg (i+3)

Assuming here arg i is 90 [although it is not possible algebraicly]
arg (i+3) must = 0?
for all i know to have an argument of 0 you need to have an arg of a real positive real number.

So please gm.haider before you guess the answers - check what you've done.

-------------------------------------------------------------

And mark - you're not right.

(x + 1.5)^2/2.25 - y^2/2.25 = 1 is just the same as people said before but changed.

 

[kooltrainer]

OMG.. thats easy but i got it wrong.. This question was in my complex test paper and i stuffed it up.. now i no how to do it.. all u do is draw the locus of that, then find its center using trig .. and u'll end up with an equation of circle if u find out its center/radius ..
easy!

 

[YannY]

No mate.. what you did was a question with the arguments subtracting one another. This is where you add them. The answer is not a circle it is a hyperbola - unfortunately for you, you still dont know how to do it even if it was in your test :mad1: .

 

[roadrage75]

This attempt is almost the same as my last, except with one small change.

we must appreciate that z lies in teh first quadrant. - if it were to lie in the second quadrant, at least one of teh angles would be obtuse - hence the sum of the angles could not be pi/2. If it were to lie in the third/fourth quadrants, both angles would be negative, and hence the sum could not be pi/2.

hence: the restrictions to teh curve are: x,y >0

Arg z(z+3) = pi/2

so arg(x^2 -y^2 +3x + 5xiy) = pi/2

since tan pi/2 is undefined, we want x^2 -y^2 +3x to = 0

we also must appreciate that x cannot = 0 and y cannot = 0

finally, we must apprecaite that since the argument is positive pi/2, we want y and x to be >0

hence the answer is x^2 -y^2 +3x =0, where y>0, x>0

hope this is right

 

[Yip]

lol I wonder why no one tried brute forcing it XD
Here is an alternate brute force approach:
arg(z)+arg(z+3)=pi/2
arctan(y/x)+arctan[y/(x+3)]=pi/2
Take the cosine of both sides (note that after cosining both sides, the trigonometric addition formula cos(x+y)=cosxcosy-sinxsiny must be used)
[x(x+3)]/[(x^2+y^2)[(x+3)^2+y^2]]=y^2/[(x^2+y^2)[(x+3)^2+y^2]]
x^2+3x=y^2, which is a hyperbola.
The only way to get the arctans to add to positive pi/2 is when y/x>0 and y/(x+3)>0, i.e. x>0 and y>0, as previously stated by the above poster. x=y=0 is invalid as an argument of a point is undefined.
Hence the cartesian equation of Arg z(z+3) = pi/2 is x^2-y^2+3x=0, with x>0 and y>0.
This is prolly not as elegant as the previous methods posted, but it requires the least thinking imo XD just a plug and chug lol

 

[A High Way Man]

what? arent we trying to FIND x and y, the coefficients of the complex number z?

 

[undalay]

two negative obtuse angles also can make an arg of pi/2, so doesn't need to be restricted to the first quadrant.

 

[GaDaMIt]

I dont believe you made it up

Either that or I dont want your answer as I do not trust in your mathematical abilities

WHOS WITH ME PEOPLE!!!!!!!!!!!!!!!!!!!11111111!!oneone1

Christ im bored

First night im not doing anything in a long while

 

[YannY]

Haha you dont have to trust my answer nor do you have to believe what i say dear entity. However, i would like to tell you that you should believe me in order to save yourself searching for a question like that which is a wild goose chase. =] Btw who knows... what if one day all the maths we did was just nonsense and useless. so believe what you like, you might be the right one. Good luck though. hehe

 

[YannY]

Oh since its christmas... hehe heres the answer x^2-y^2+3x=0 within the region 2xy + 3y > 0. As you can see, mark initially got this right, however he simplified the inequalities wrong... he gave the final inequality of x>-3/2 Which is not really the simplfied version of 2xy + 3y > 0.

hmm interesting aint it. Anyways i know many of you would think wtf? this guy is nuts why isnt that the simplified version. well good question but it also raises the question of whether you should relearn your inequalities or not. Maybe not, i was jk you dont have to relearn it, this inequality is quite the complicated one to be in the hsc. All you'll need to know is the algebra of harder one dimension inequalities + the graphical drawing of simple two dimension inequalities.

Anyways if you dont believe that the inequality cant be simplified into that just x>-3/2. I shall give you a simple proof~[damn my number 1 button aint working hence no exclamation mark] test the point [-2,-1] into the equations. For those lazy people here it is for both eqN.
For 2xy + 3y > 0 -> 2[-2][-1]+3[-1]>0
= 4-3>0
= 1>0

Is this true? well duh 1 is obviously greater than 0.

But if we test that exact same point into marks given region x>-3/2

-2>-3/2

If thats true in your world mark, im sorry i bothered you, but i wonder if you can lend me the portal to your world, it seems fun there.

Anyways if you still dont believe me, merry christmas. Do not proceed.

OKAY for an alternative answer, we know that in the second quadrant the answers will turn obtuse if added[as said before], in the first quadrant the answer is correct hence let the angles be A and B[in the first quadrant] hence A+B=90. in the second quadrant the angles will be 180-A and 180-B. Hence when they add it will be 180-A+180-B=360-[A+B]
= 360-90
=-pi/2 in the domain of -pi<x<pi
So it isnt in the 2nd quadrant.

In the 4th quadrant it will be -A+-B=-90
=-pi/2

Hence that is not true either.

For the 3rd quadrant it will be A-180 + B-180 = A + B - 360
= 90 - 360
= -270
=pi/2

Hence the hyperbole curve is only correct in the 3rd and 1st quadrant not including the axis.

Remember to get this, we must establish that A+B=pi/2 is true in the first qudrant.

It is recommended that you do the question in the second method as oppose to marks method. Why? easy, because harder inequalities with 2+variables are long and tedious, its takes the same time as doing two of these questions and plus its errorneous. The second method is boring, yes it does look boring and it took me long to type, thats cause my keyboard is broken. Writing this on paper takes 5seconds. Also it is safe, and easily understood by teachers. I mean what if you did wrote 2xy + 3y > 0 down but teacher didnt understand and simplified like mark and gave you the wrong mark, haha there goes another pun. Excuse me.

Also you may use the brute forcing method but you must test the qudrants.
For those who may not know what arctan is, its just inverse tan.

Of course, there are many other ways to do this, but follow the second method and thats all you'll need. If anyone else finds another method please post to show your intelligence.

Merry christmas, im going to melbourne~ yay.

 

[YannY]

what? arent we trying to FIND x and y, the coefficients of the complex number z?

No this is a locus question... Definition of Locus: Mathematics. the set of all points, lines, or surfaces that satisfy a given requirement.
http://dictionary.reference.com/browse/locus

Hehe x+y, yeah sure you can find that, but you'll need to give an infinite answer that way, good luck =]. And if you can, you'll be my hero.

 

[sXcGuRl69aU]

hahahahahahaha fuck

 

[YannY]

Dude, it's Christmas... go be with your family...

My family is this computer and my maths textbooks man... i am with my family in other words. Plus bored of studies is my family too. What about you? is your family BoS as well? LAWL honestly when you say something like that you should really plan it... Im asian, we dont celebrate christmas[all we do is wish people christmas but we dont mean it really] just like you dont celebrate moon festival or sushi day and if you did... G4U and dont do it again, its not a festival you are allowed to celebrate unless you're asian. Haha im jk you can celebrate it but i dont have time to celebrate all the festivals that are.

Besides if you read the whole post you'd know i promised to give out the solutions on christmas. Lawl merry christmas dear entity who celebrates christmas with his computer. hohoho

 

[Slidey]

This dead horse has been beaten before.

http://community.boredofstudies.org...n-2/67161/complex-no-q.html?highlight=complex

Emotions flew like eagles in that thread, too.

 

[Undermyskin]

Is the answer a circle centre (0,-1.5) radius 1.5? and the point (0,-3) and the origin are two heads of the diametre? z is different from 0 and -3??? Maybe not the whole circle tho???

 

[samwell]

Argz(z+3)=pi/2
z(z+3)=o
z=0+0i ang -3+0i
see the diagram in attachment.

 

[YannY]

you still trying? well thats not the answer mate and you're far from it.

 

[Slidey]

you still trying? well thats not the answer mate and you're far from it.

That came off overly negative. I hope you're not implying that posters shouldn't attempt previously solved questions? Because that would be a very bad idea.

And if a poster isn't correct, please give constructive criticism instead of telling them they're completely wrong.

 

[YannY]

Sorry slidey and sorry preposters

Nice try - Hint: argument of z is usually a locus not sets of points.

But hey nice try - keep trying.

 

[adiba90]

hi! hey i know this is pretty pathetic of me to like ask you for help on this. Well to be askign help off a stranger even. But yeah im sorta stuck. Could you please try and help me. even direct me

okay so i need help with Algebra. the question is can i simplify >>>>

arccos(sin(5pi/4)) ??? man this would help me alot if you know it! Thanks!