Argand Diagram (1 Viewer)

[nrlwinner]

How do you draw arg(z-a)=arg(ia)

At first I thought it was the 2 lines extendng from points a and ia, but then I realised arg(ia) isn't the same as arg(z-ia)

 

[shaon0]

How do you draw arg(z-a)=arg(ia)

At first I thought it was the 2 lines extendng from points a and ia, but then I realised arg(ia) isn't the same as arg(z-ia)

arg(z-a)=pi/2+arg(a)
arg(z-a)-arg(a)=pi/2

Draw a parallelogram then construct a diagonal through z and a. arg(z-a)-arg(a) is the angle from (z-a) to a and this has to be equal to pi/2.

 

[_henwee]

Question... Can calculators graph Argand Diagrams?

 

[study-freak]

How do you draw arg(z-a)=arg(ia)

At first I thought it was the 2 lines extendng from points a and ia, but then I realised arg(ia) isn't the same as arg(z-ia)

arg(z-a)=arg(ia)
arg(z-a)=pi/2+arg(a)
arg(z-a)-arg(a)=pi/2

Re(z)=a, Im(z)>0 if a>0
Re(z)=a, Im(z)<0 if a<0
assuming that a is a real number


OR if a is not necessarily real,
the locus will be a tangent of the circle x^2+y^2=a^2
with restrictions:
Im(z)>Im(a) if a is in the 1st or 4th quadrant (excluding imaginary axis)
or Im(z)<Im(a) if a is in the 2nd or 3rd quadrant (excluding imaginary axis)
or Re(z)<0 if a lies on the positive imaginary axis
or Re(z)>0 if a lies on the negative imaginary axis.

 

[shaon0]

Question... Can calculators graph Argand Diagrams?

Try graphing calculators just enter arbitrary lines like y=2x or y=(1/2)x and let them be z and a

 

[nrlwinner]

How do you draw x^2-y^2

 

[study-freak]

How do you draw x^2-y^2

You mean x^2-y^2=a^2?

 

[nrlwinner]

You mean x^2-y^2=a^2?

Yeah something like that where x^2-y^2=constant

 

[study-freak]

Yeah something like that where x^2-y^2=constant

Well, x^2-y^2=a^2 is a rectangular hyperbola that goes through (a,0) and (-a,0).
It has asymptotes y=+-x.
For details, read conics section of your MX2 textbook.

 

[nrlwinner]

Well, x^2-y^2=a^2 is a rectangular hyperbola that goes through (a,0) and (-a,0).
It has asymptotes y=+-x.
For details, read conics section of your MX2 textbook.

Gotcha. Haven't done conics that's why it's new to me.
Also, how do you draw the ellipse with complex number |z-z1|+|z-z2|=2a

 

[study-freak]

Gotcha. Haven't done conics that's why it's new to me.
Also, how do you draw the ellipse with complex number |z-z1|+|z-z2|=2a

You need to use the fact that:
PS+PS'=e(PM+PM)=eMM'=e(2a/e)=2a.
S and S' are foci of the ellipse.
So we let z1 and z2 be the two foci and z be represented by P.

Now that equation above makes sense because PS+PS'=2a.
a=length of semi-major axis

And we know that the distance from z1 to z2=2ae (distance between two foci)
e=eccentricity

We can now work out a and e, and using b^2=a^2(1-e^2)
we can also work out b.

Now, you can draw it.

 

[nrlwinner]

You need to use the fact that:
PS+PS'=e(PM+PM)=eMM'=e(2a/e)=2a.
S and S' are foci of the ellipse.
So we let z1 and z2 be the two foci and z be represented by P.

Now that equation above makes sense because PS+PS'=2a.
a=length of semi-major axis

And we know that the distance from z1 to z2=2ae (distance between two foci)
e=eccentricity

We can now work out a and e, and using b^2=a^2(1-e^2)
we can also work out b.

Now, you can draw it.

Thanks. One last thing.
If I was to find the locus of z^2 where z=x+iy, what would the graph look like?

 

[study-freak]

Thanks. One last thing.
If I was to find the locus of z^2 where z=x+iy, what would the graph look like?

What's z?
We can never draw something like that without defining what z is.
z=x+iy is not a definition since x and y are variables.

 

[nrlwinner]

What's z?
We can never draw something like that without defining what z is.
z=x+iy is not a definition since x and y are variables.

The question was Re(z^2)+Im(z^2)=0

 

[study-freak]

The question was Re(z^2)+Im(z^2)=0

z=x+iy
z^2=(x^2-y^2)+(2xy)i

Re(z^2)+Im(z^2)
=x^2-y^2+2xy
=0

Now, we've got the equation and we need to implicitly differentiate, etc to draw it (graphs chapter).

 

[nrlwinner]

z=x+iy
z^2=(x^2-y^2)+(2xy)i

Re(z^2)+Im(z^2)
=x^2-y^2+2xy
=0

Now, we've got the equation and we need to implicitly differentiate, etc to draw it (graphs chapter).

Wait. I thought you can't do implicit differentiation when the constant is 0 because when solving for vertical and horizontal tangents, it will keep coming out as 0.

 

[shaon0]

The question was Re(z^2)+Im(z^2)=0

Re(z^2)+Im(z^2)=0
Re(x^2+2ixy-y^2)+Im(x^2+2ixy-y^2)=0
x^2-y^2+2xy=0
y^2-2xy-x^2=0

y=(2x+-sqrt(4x^2+4x^2))/2
=x+-xsqrt(2)
y=x(1+-sqrt(2)) ie y=x+sqrt(2).x OR y=x-sqrt(2).x

Implicitly differenitiating i get: turning pt and a vertical asymptote at (0,0)

 

[study-freak]

Re(z^2)+Im(z^2)=0
Re(x^2+2ixy-y^2)+Im(x^2+2ixy-y^2)=0
x^2-y^2+2xy=0
y^2-2xy-x^2=0

y=(2x+-sqrt(4x^2+4x^2))/2
=x+-xsqrt(2)
y=x(1+-sqrt(2)) ie y=x+sqrt(2).x OR y=x-sqrt(2).x

Implicitly differenitiating i get: turning pt and a vertical asymptote at (0,0)

Yeah, I get the same.
And so it doesn't make sense lol.
turning point and vertical tangent at the same point?

 

[shaon0]

Yeah, I get the same.
And so it doesn't make sense lol.
turning point and vertical tangent at the same point?

Maybe solving as I did is the method to go but doesn't seem right. Additionally i tried: (x/y)-(y/x)=-2 which was my first thought.

(x^2+y^2)(xy'-y)=0 ...maybe x=+-iy but same problem as before

 

[nrlwinner]

Yeah, I get the same.
And so it doesn't make sense lol.
turning point and vertical tangent at the same point?

Yeah. That's why I've never been able to implicitly differentiate anything when the constant is 0.