anonymous
  • anonymous
Arg(z+i)=0 Given this information im supposed to draw the relevant graph
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I got it down to (y+1)/x=0 or pi but im not sure if i went about it right
anonymous
  • anonymous
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)
amriju
  • amriju
If its Arg not arg.it may refer to principle values only..in this case 0..what u've done is correct...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.