How would you sketch this? (1 Viewer)

[HyperComplexxx]

 

[deterministic]

So you are really sketching:
This will basically be the first quadrant rotated by clockwise

 

[HyperComplexxx]

oh yea ...
so simple

 

[blackops23]

So you are really sketching:
This will basically be the first quadrant rotated by clockwise

noice

 

[ZachBC_94]

wouldn't:

???

 

[xV1P3R]

I think you're mistaking it with arg[a] + arg = arg[ab]

So it's the other way around (products and sum)

 

[khorne]

It would work the same Arg(z +iz), but arg(z+iz) does not expand to arg(z) + arg(iz)

 

[kr73114]

is that similar to arg(z-2/z+2)=pi/2
how would u do that?

 

[Patato]

arg(z-2/z+2) = pi/2 is a semi circle with centre 0,0 and radius 2 found using either circle geometry rules

 

[Drongoski]

arg(z-2/z+2) = pi/2 is a semi circle with centre 0,0 and radius 2 found using either circle geometry rules

Is it then the upper semi-circle or lower or both?

 

[kr73114]

arg(z-2/z+2) = pi/2 is a semi circle with centre 0,0 and radius 2 found using either circle geometry rules

how do u know?

 

[Drongoski]

how do u know?

You can draw the circle centre origin radius 2. You take a convenient typical point z in the upper semicircle and see if arg(z-2) - arg(z+2) = pi/2 or - pi/2. Do the same for convenient typical point z on lower semi-circle.

 

[Patato]

you can break up arg(z-2/z+2) into arg(z-2) - arg(z+2), so sketch two vectors that intercept to make a right angle, (their args subtract to give pi/2), and you know from circle geometry that an angle at the circumference of a circle when subtended by the diameter of the circle will be a right angle, therefore you know its a circle, with its centre half way between the origin of the two intersecting vectors...so 0,0. but its a semi circle because if the vectors went into the negative half of the y axis, the arg would be negative if we are talking about the primary argument. id sketch it for you but im on a laptop and i cant stand the touch pad in paint

 

[cutemouse]

You can draw the circle cente origin radius 2. You take a convenient typical point z in the upper semicircle and see if arg(z-2) - arg(z+2) = pi/2 or - pi/2. Do the same for convenient typical point z on lower semi-circle.

Or deduce it from the external angle property of a triangle... (easier, I think).

 

[blackops23]

Or deduce it from the external angle property of a triangle... (easier, I think).

yep, this is the way i do it, by FAR the easiest... if only I had a diagram...

 

[khorne]

yep, this is the way i do it, by FAR the easiest... if only I had a diagram...

remember 1 is a solution cause (1)arg(z-2/z+2) = pi/2

 

[ZachBC_94]

Don't forget 2 and 1/2, seeing as how (2)(1/2)(1)arg(z-2/z+2) = pi/2

 

[khorne]

stfu m8, you don't get the joke