truly Complex. (1 Viewer)

[YannY]

[Q. Given |z|=1 find the locus of z^2.]

I look at this q and its wtf BBQ, it dosnt make sense to me because z^2 has an imaginary part and i honestly dont know how to plot a locus with an imaginary part.

First attempt:
z^2=x^2-y^2+2xyi
x^2+y^2=1

Thats about it...

Later in the day -

I tried using mod-arg form, this is what i get
z=cis@
z^2=cis2@

So i concluded this would be a dotted circle in the complex region with centre at nought. But then is this really a circle? I asked myself.

Somebody help me?

 

[undalay]

This is how i approached it.

|z| = 1
|Z^2| = 1

arg (z) = a
arg(z^2) = arg(z) +arg(z) = 2a
thus arg(z^2) can be all arg.


i.e. unit circle?

 

[conics2008]

|z| = root of x^2+y^2

hence (root of x^2+y^2)^2=1

x^2+y^2=1 which is a circle with center (0,0) and radius 1

 

[YannY]

This is how i approached it.

|z| = 1
|Z^2| = 1

arg (z) = a
arg(z^2) = arg(z) +arg(z) = 2a
thus arg(z^2) can be all arg.


i.e. unit circle?

Thought of that, makes no sense.

For example: instead of asking find the locus of z^2, i ask find the locus of z..

Now i can say
|z|=1
arg z = a
Like you: thus arg(z) can be all arg.

Hence z is a unit circle.

No this is not true because a locus cannot exist on the complex region, z is simply a complex number with modulus of one.

Tell me im wrong lol... im starting to consider this question being faulty.

 

[YannY]

|z| = root of x^2+y^2

hence (root of x^2+y^2)^2=1

x^2+y^2=1 which is a circle with center (0,0) and radius 1

Thanks for trying but i didnt ask find the locus of |z|^2, the q is find the locus of z^2 which is (x+iy)^2.

 

[undalay]

I dont understand ur explanation.

The locus of z is the unit circle?
Do you agree?

Now if u square any vector on this circle...its still going to lie on the unit circle!

Simply because the squared vector's modulus is gonig to be 1^2 = 1, and the arg is going to be 2x the arg of z, and since the arg of z can be anything, the arg of z^2 can be anything.

wat doesn't make sense?

edit: conic's worknig out looks fine to me also

 

[Iruka]

It's a locus in the Argand plane. Complex number questions just about always have a geometric interpretation as a complex number can always be thought of as a 2d vector of real numbers.

If you think of it as a function of theta, where theta = arg(z), then z^2 just travels around the unit circle twice as fast as z.

 

[YannY]

no such thing as a locus on complex region

 

[YannY]

if there is give me a clear example and explain

 

[hon1hon2hon3]

I dont see what dosent make sense from explaination above . .

Since lZl = 1 <- for this, this is a unit circle with radius 1

and Since lZl=1 then therefore lZ^2l is also equal to 1 ... lZ^2l=1

From lZl = 1 <- a unit cricle . . graph it , its argument can be all arg.

and for lZ^2l=1 we can say that 2 arg z , but we already know that arg z can be all arg . . . so whats is two times that ? its still all arg

So my the way i do it , the locus of lZ^2l = 1 , its the same graph with lZl = 1.

Just that arg Z^2 = 2 arg z .

I am noobish in complex number , if i have explained anything wrong , please correct me . Peace

 

[Managore]

if there is give me a clear example

Umm... The locus of z if |z| = 1?

Ask yourself "What [complex] values satisfy |z| = 1?". The answer is of course the unit circle. Since the question is obviously dealing with the complex plane, the locus will be all complex numbers which satisfy |z| = 1. If the question is asking about points on the 2D plane, the locus will include all appropriate points on the 2D plane.

A locus is simply a set of points which satisfy the given conditions. The domain which you are dealing with may be the real line (R), or the 2D plane (R^2), or the complex numbers (C), or any domain you can think of.