Complex numbers (1 Viewer)

[Premus]

Sketch

im z = /z/

when i sub in z = x + iy , i get x = 0

but the answers say x= 0 for y >= 0
am i missing something silly?


Thanks

 

[nit]

/z/ >= 0 as it is a distance. Hence y is also >= 0

 

[BillyMak]

nit ---> Target UAI 100
Is this a serious, realistic target?

 

[CrashOveride]

y = |z|
x² = 0
Hence x =0, for y>=0 because of the square root up there, and as nit pointed out its a distance. This was a HSC question i feel.

 

[Premus]

oh alright
Thanks!

 

[ressul]

What's the answer?

So the graph would be liked a line? That from 0 to infinity?

 

[CrashOveride]

Starting from the origin and going up the imaginary axis, so you just put an arrow indicating its continuous.

 

[LSP]

how to find the locus of z

arg(Z+i) = arg (Z-i)

 

[mr EaZy]

Starting from the origin and going up the imaginary axis, so you just put an arrow indicating its continuous.

thats right, also indicate that it ends at (0,0) and has an arrow sign, i dont think thay care about the arrow though.

 

[Archman]

how to find the locus of z

arg(Z+i) = arg (Z-i)

Z = yi where y>1 or y< -1

 

[mr EaZy]

was that a question? ohh.

i thought it would have been Z=yi but Y>-1 for some reason
i expected both to have args pi/2
am i missing something here?

 

[Premus]

how did u do that question?
arg(Z+i) = arg (Z-i)

Thanks

 

[Estel]

For arg(z+i) = arg(z-i), z must lie on the y axis. (use diagram to convince yourself)
:. z = iy
but if -1<y<1, the arguments are going to be opposite in sign. So z has to lie either below both i and -i, or above both i and -i.
so y<-1, y>1

Same situation with arg(z-1)=arg(z+1)
z has to lie on the real axis and be to the left of both -1 and 1 or to the right of them both.

 

[Veck]

how did u do that question?
arg(Z+i) = arg (Z-i)

Thanks

since I don't understand what people have done to answer this question, I'll post this:

arg (z + i) = arg (z - i)
arg (x + i (y+1)) = arg (z + i (y-1))
tan^-1 ((y+1)/x) = tan^-1((y-1)/x) [take tan of both sides]

(y+1)/x = (y-1)/x
xy + x = xy - x
2x = 0
x = 0

hmm.. again the same answer.
yeah cool.

 

[LSP]

thanks alot