Question

Arg(z+i)=0
Given this information im supposed to draw the relevant graph

Answer 1

I got it down to (y+1)/x=0 or pi but im not sure if i went about it right

Answer 2

Ill go through my process if it helps anyone viewing
Arg(z+i) = Arg(x+iy+i)
therefore Tan^-1((y+1)/x) (definition of Arg) = 0
and then (y+1)/x=tan(0)

Answer 3

If its Arg not arg.it may refer to principle values only..in this case 0..what u've done is correct...

Answer 4

r u sure all values can be taken...of k??..@CarlosGP ...nd not just 0?

Answer 5

what Taplin did is correct...he just has to choose the plot for the principle value of Atan (0)

Answer 6

I made a mistake and solved for Tan (y+1/x) instead of Atan(y+1/x), his solution is right!

Answer 7

Yes it is Arg, so what your saying is that it cannot be pi because the principal domain of tan is (-pi/2 to pi/2)
and hence my eqaution is just y=-1... ?

Answer 8

you are right

Answer 9

Much simpler is that if Arg(z+i)=0--->Im(z+i)=0--->Im(x+i(y+1))=0 ---->y+1=0 ->>>y=-1

Answer 10

well...standard books refer to Arg when only principle value is considered...arg when general values r considered..nd the domain is -pi to pi I think...

Answer 11

Ah, thankyou both very much, been a while. If i was to sketch it it would only occur in the 1st and 4th quadrants correct. Im pretty sure the principal of tan is pi/2 though. unless im sorely mistaken

Answer 12

let me check...

Answer 13

btw @CarlosGP i dont quite understand your reasoning behind if Arg(z+i)=0--->Im(z+i)=0

Answer 14

u check the net or any book..u'll see i'm correct....(unless they take it from 0 to 2pi that is also sometyms taken)...u have to have a 2pi interval...check it..

Answer 15

'By convention it is the interval –π/2≤ x≤ π/2 for functions sin(x) and tan(x)'

Answer 16

@Taplin44 If Arg(z+i)=0 then z+i has to be a Real Number, which means its imaginary part is zero---->Im(z+i)=0--->y+1=0--->y=-1

Answer 17

R u checking it in trigonometry??...check for priciple value in case of arguments of comp.numbers

Answer 18

@Taplin44 got it now?

Answer 19

ahhh yes, i see, makes sense now, thankyou !
And you are indeed correct 'The principal value of tan−1θ is always between −π2 and π2. The principal value of argz, on the other hand, is always in the interval (−π,π]'
Would i not then have a second possible equation of y=xpi-1 though if that is the case ?
sorry for being a pain i just really wanna understand this for my exam

Answer 20

I made a mistake with y=k*pi-1, forget about it. The solution is y=-1, that is a horizontal line crossing y-axis in y=-1

Answer 21

And x takes values from -infinity to +infinity

Answer 22

That all makes sense, except for the fact that if pi is in principal interval then shouldnt (y+1)/x = pi be a possible solution ?

Answer 23

No, you are making the same mistake I did. If Arg(z+i)=0--->Atn(y+1/x)=0 and that means that y+1/x=0---->y+1=0---->y=-1

Answer 24

ohhhhh yes, thankyou ! very very helpful