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I got it down to (y+1)/x=0 or pi but im not sure if i went about it right
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)
If its Arg not arg.it may refer to principle values only..in this case 0..what u've done is correct...
r u sure all values can be taken...of k??..@CarlosGP ...nd not just 0?
what Taplin did is correct...he just has to choose the plot for the principle value of Atan (0)
I made a mistake and solved for Tan (y+1/x) instead of Atan(y+1/x), his solution is right!
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... ?
you are right
Much simpler is that if Arg(z+i)=0--->Im(z+i)=0--->Im(x+i(y+1))=0 ---->y+1=0 ->>>y=-1
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...
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
let me check...
btw @CarlosGP i dont quite understand your reasoning behind if Arg(z+i)=0--->Im(z+i)=0
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..
'By convention it is the interval –π/2≤ x≤ π/2 for functions sin(x) and tan(x)'
@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
R u checking it in trigonometry??...check for priciple value in case of arguments of comp.numbers
@Taplin44 got it now?
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
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
And x takes values from -infinity to +infinity
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 ?
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
ohhhhh yes, thankyou ! very very helpful