graphing complex numbers help!!! (1 Viewer)

[poptarts12345]

how do I do this question??

Sketch the graph of:
(a) arg(z + i) = arg(z − 1)
(b) arg(z + i) = arg(z − 1) + π



ANSWERS :
a)
b)

 

[tito981]

when you have a argz equation, the general form for graphing the points is arg(z-(x+iy)), where x+iy is the point where you have the start of the ray. When there is 2 args in an equation arg(z)=arg(w), you see where they overlap. e.g. for (a) the equation is arg(z-(0+i))=arg(z-(1+0i)), so you plot the points, 0+i, 1+0i, and see when the angles are the same which is when they pass through each others points. The locus cant be in between the points as they will have args which differ by pi.

for (b), the pi is a rotation of 180 deg anti-clockwise. so if the arg for both sides is pi/4, the addition of the pi, makes it 5pi/4 (-3pi/4) which is the line segment between the points. hope this helped.

 

[CM_Tutor]

how do I do this question??

Sketch the graph of:
(a) arg(z + i) = arg(z − 1)
(b) arg(z + i) = arg(z − 1) + π

An easy way to think of this is to recognise that refers to the vector from to .

So, means that the direction from to is the same as the direction from 1 to .

Let A correspond to the point where and B correspond to the point . For any point Z (corresponding to the complex number ) not on the line through A and B, ABZ would be a triangle with three non-zero angles. It follows that the directions of vectors AZ and BZ cannot be the same, and thus that all points in the locus must lie on the line through A and B... though not all points are necessarily included.

The points A and B themselves cannot be in the locus as, in each case, one of the required arguments is defined and the other is not because
is undefined.

The points between A and B (where ) are also excluded from the locus as in this region



but



In other words, between A and B



and so the interval AB (but excluding its end points) is the locus for part (ii), .

For all points on AB with , the required arguments are the same because the directions of the vectors AZ and BZ are the same, and so those points are included in the locus of .

 

[beetree1]

LIFESAVING TIP:


a) arg(z + i) = arg(z − 1)

• Rearrange this and you will get arg(z + i) - arg(z − 1) = 0 --> Which is the same as arg(z - (-i)) - arg(z − (1)) = 0
• From here, you can plot the points (0,-1) and (1,0).
• Now on the RHS of your equation, you have the number zero - note the bolded O. You can think of this as outside. So thus, you would sketch the outside of the line joining the points (0,-1) and (1,0), and not the inside.

(b) arg(z + i) = arg(z − 1) + π

• Similarly to above, rearrange this and you will get arg(z + i) - arg(z − 1) = π --> Which is the same as arg(z - (-i)) - arg(z − (1)) = π
• From here, you can plot the points (0,-1) and (1,0).
• Now on the RHS of your equation, you have the number pi - note the bolded i. You can think of this as inside. So thus, you would sketch the inside of the line joining the points (0,-1) and (1,0), and not the outside.