hard but not impossible complex q. (1 Viewer)

[YannY]

Okay ive a very hard question of arguments. haha scuse the pun.

find the cartesian equation of Arg z(z+3) = p/2

Dont ask where i got this question cause i made it up.

PS: dont say its impossible. if you do i'll hunt you down for giving up on maths questions when you're doing 4u. just for your comfort, its only 0.001% that it would appear in your test. For those who are doing 4u maths without any hopes... 0.001% is not 0.001 of a thing its 0.00001.

who ever can do it deserves a cheer from us.

Merry christmas, i'll give you the solutions maybe next year...

 

[tommykins]

x = -3. But only in the positive side of the y axis (circle at x = -3, pointing straight up)

Arg z(z+3) = p/2

arg z (z-(-3+0i)) = pi/2.

point is (-3,0) and since the argument for z is pi/2 it is a straight line perpincular to the x axis at -3.

Unless the question is the arg of z(z+3) = z^2 + 3z?

if so, I'll get back to you.

 

[gm.haider3000]

just quickie z=-3i havnt got up to it yet still up to 2.3 vectors

 

[Trebla]

This is assuming you mean find the cartesian equation of arg[z(z+3)] = π/2
Let z = x + iy where x and y are non-zero and real
z(z + 3) = z² + 3z
= (x + iy)² + 3z
= x² - y² + 2ixy + 3x + 3iy
= x² - y² + 3x + i(2xy + 3y)

But if the argument of this is π/2 then it must be purely imaginary. This is because if you let z(z + 3) take the mod-arg form r.cis θ, where θ = arg[z(z+3)] and r is the modulus of z(z+3), when θ = π/2, you're left with i.r which is purely imaginary.

Hence the real part must equal zero; Re[z(z + 3)] = 0
i.e. cartesian equation is x² - y² + 3x = 0 => y² = x² + 3x
Note that x and y are non-zero, because if x = 0, then y = 0 to satisfy the locus and consequently you get z = 0 which follows that arg (0) = π/2. The argument of the origin should be undefined.

I'm not sure if that's correct, there might be a flaw somewhere, but it makes sense in a way.....

 

[tommykins]

Hence the real part must equal zero; Re[z(z + 3)] = 0
i.e. cartesian equation is x² - y² + 3x = 0 => y² = x² + 3x
Note that x and y are non-zero, because if x = 0, then y = 0 to satisfy the locus and consequently you get z = 0 which follows that arg (0) = π/2. The argument of the origin should be undefined.

I'm not sure if that's correct, there might be a flaw somewhere, but it makes sense in a way.....

I got up to that and had no idea how to draw it. =D

just quickie z=-3i havnt got up to it yet still up to 2.3 vectors

Wouldn't that make a horizontal line at -3 on the Y axis? making the argument pi?

 

[gm.haider3000]

tommy not really its the y value it would give 90 the arg.. yeah my first attempt was let z = z+iy then i lost hope soo i said ill just guess cuz the arg was 90 but when u draw z=-3i its not pi its pi/2 actually negative pi/2.....

 

[YannY]

No the question is correctly arg z(z+3) = pi/2

and i specified pi/2 because that will keep you guys thinking, if it was like pi/4 then even my 3unit buddy can probably do it.

lemme hint for those who've got no idea or people who are trying out the question but isnt doing 4u confused: ) lol.

[arg z(z+3) = pi/2] is the same as:
arg z + arg (z+3) = pi/2

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

Trebla dont give up.. you're nearly there. oh yeah, for those who've got tutors go ahead and ask them... if they cant answer it then get a new one- obviously they dont understand maths (you dont want a dota player to teach you how to play soccer do you?).

Anyways i wont give the answer just yet but i will after christmas if no one gets it - so its a good present for you all. :rofl: .

Good luck... may the theorems be with you.

 

[tommykins]

tommy not really its the y value it would give 90 the arg.. yeah my first attempt was let z = z+iy then i lost hope soo i said ill just guess cuz the arg was 90 but when u draw z=-3i its not pi its pi/2 actually negative pi/2.....

Ahh of course, i read it as y = 3i lolz.

apologies.

 

[YannY]

btw its not z = 3i either. if you sub it in, it will give :

LHS: arg(3i) + arg (3+3i)
= undefined + pi/4

unless arg (3i) is pi/4 then you are right.

 

[roadrage75]

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 to be >0

hence the answer is x^2 -y^2 +3x =0, where y>0, and x and y cannot = 0, and x cannot = -3

hope this is right

 

[YannY]

Good work... again.. like trebla you're nearly there the y>0 part which you added is only partially correct. =( nice try.

BTW you dont need to say both y and x cant be 0. this is because when x = 0 y is also 0. Hence by saying x ¹ -3, 0 would have been correct. Correctly that would actually give: Z ¹ -3, 0.

Good luck, keep trying.

 

[A High Way Man]

Ive never seen a question like this at Uni.

Do you mean: Find the Cartesian equation of z, given that Arg(z*(z+3)) = Pi/2

 

[YannY]

yes that is what i mean and i posted it clearly two times already. its quite obviously since its multiplication there should be no need for the extra parenthesis outside.

 

[gm.haider3000]

yanny i dont think u can get tan 90 its undefiend ?? unless its something else becuase your arg= tan y/x ?? hmm still cant seem to figure it out can u like give us the mod

 

[Mark576]

Well as has been mentioned above, the argument is positive pi/2, and accordingly, y > 0, and as we know that if the Arg Z = +- (pi/2) gives a purely imaginary complex number Z, then there must be some reason why it doesn't equal - (pi/2). So then Im (z) > 0. From this we see that 2xy + 3y > 0, and since y > 0, this then simplifies to x > -3/2. From observing the graph drawn of x^2 - y^2 + 3x = 0, we see that the cartesian equation must be x^2 - y^2 + 3x = 0, where x > 0, as it has already been said that x =/= 0. Ergh, not really sure about this aye

 

[gm.haider3000]

one more thing is it Argz = z+3 or z(z+3) ??? im getting even more confused because no way in the world can an argument be 90 degree unless it only has an imaginary number ?
??? soo it will help if you give us the mod

 

[gm.haider3000]

mark i did that method where you y cancels out but then it does not staisfy both of your equation which is x^2-y^2+3x and 2xy+3y=0

 

[gm.haider3000]

lol did rearrangin hahah
maybe there could be two complex numbers
haha like division could be a possiblity
or to complex numbers multiplying each other eg one could have arg 60 and the other arg of 30 then when added can give u a arg of 90 which doesnt have to be a straight line ill try it now

 

[undalay]

Arg z(z+3) = pi/2

let z = x+iy

z(z+3)
= (x+iy)(x+iy+3)
= x^2 + xiy + 3x + xiy -y^2 +3iy
= x^2 +3x - y^2 + i(2xy + 3y)

arccot( (x^2 +3x - y^2)/ (2xy + 3y)) = pi/2
Cot both sides

x^2 +3x - y^2)/ (2xy + 3y)) = 0

x^2 + 3x - y^2 = 0

(x + 1.5)^2 - y^2 = 2.25

(x + 1.5)^2/2.25 - y^2/2.25 = 1

 

[gm.haider3000]

hey yanny is the equation z=i