Transformations of the complex plane

gauss_is_back

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Could somebody please give me some help for the following homework problem.
Does anyone know a technique for these sorts of questions?
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The complex numbers z and w are represented by the points P(x, y) and Q(u, v) respectively in Argand diagrams and w = z2

The point P moves along the curve with equation 2xy2 = 1.

Find the equation of the locus of Q, giving your answer in the form u = f(v).

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Also, if you have time (your help would be TREMENDOUSLY appreciated!!) here's another question:

The points P(x, y) and Q(u, v) represent, respectively, the complex numbers z and w. Given that w = z2 find the equation of the locus of Q when the point P moves around the circle x2 + y2 = 4.
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Thank you in advance.
 
The complex numbers z and w are represented by the points P(x, y) and Q(u, v) respectively in Argand diagrams and w = z2

The point P moves along the curve with equation 2xy2 = 1.

Find the equation of the locus of Q, giving your answer in the form u = f(v).
What have you done, so far? Where did you get stuck?

We have:

z = x + y·i

w = u + v·i

where w = z^2

They want u expressed as a function of v.

In the relationship between x and y (given by the equation 2xy^2=1), y is not a function of x, but x is a function of y. So, I would begin by finding that function.

If you substitute the result for symbol x, in the definition for z, you can then determine an expression for w in terms of y.

Separate the Real part of w from the Imaginary part. The Real part is u, and the Imaginary part is v·i.

Compare the expressions for u and v. What operations performed on v will result in u?

Please be sure to show any work that you've tried, when you post again. Thanks. :cool:
 
Thank you for your reply mmm4444bot,
I have tried out your approach and here's where I'm at thus far.

Let z = x + iy, w = u + iv [we know that z is mapped to w by the function w = z2]
Writing x as a function of y gives x = 1/(2y2)
Therefore, w = (1/(2y2) + iy)2
Now when I eventually go on to equate real and imaginary parts I deduce that:
u = (1 + 4y6)/4y4 and v = 1/y.

After doing the step you described as "
Compare the expressions for u and v. What operations performed on v will result in u?", I notice that [since 1/y = 4y3/4y4]
to get from v to u, we must multiply v by 1/4y3 + y3.

Could you please describe to me how I now continue from here in order to get u as a function of v, without having x or y involved.

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NB I have been looking at some more of these types of questions and I have noticed a trend in that they often ask to obtain expressions for u and v in terms of x and y before going on to find the equation of the loci. In some (simpler) examples, I understand that if you have x and y in terms of u and v, it makes it easy to write u as a function of v straightaway (this is often the case when, for instance, the function that maps z to w is something like w=1/z (as opposed to w=z2 which is slightly more awkward).
Do you happen to know why they would ask for u and v in terms of x and y (instead of x and y in terms of u and v) and how this might be helpful in any way? By the way, I can't just manipulate the expressions as they often involve x2 and y2 (and sometimes even cubic or quartic) terms which make it both extremely awkward and very lengthy (consuming far too much time available to me in an exam).
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Once again I really appreciate your help!


 
… [equating] real and imaginary parts I deduce that:

u = (1 + 4y6)/(4y4)

v = 1/y
There's a sign error, in your expression for u. (When squaring x+i·y, don't forget that i gets squared, too.)

I had not combined the ratios, in my expression for u:

u = 1/(4y^4) - y^2


I notice that [since 1/y = 4y3/(4y4)] to get from v to u, we must multiply v by 1/(4y3) + y3.

Could you please describe to me how I now continue from here in order to get u as a function of v, without having x or y involved.
I didn't think about other forms of 1/y. I simply started with 1/y, and thought about how to get 1/y^4 and y^2, using properties of exponents.

We need a function whose input is 1/y and whose output is 1/(4y^4) - y^2.

Looking at the first term, we can get 1/(4y^4) by raising 1/y to the fourth power and multiplying the result by 1/4.

Looking at the second term, we can get y^2 by raising 1/y to the negative second power.

Subtract the results.

u(v) = 1/4·v^4 - v^(-2)

We can simplify.

u(v) = v^4/4 - 1/v^2

I would have used the same type of reasoning, were I to have started with this form, instead:

u = (1 - 4y^6)/(4y^4)

The resulting u(v) would be equivalent.
 
Do you happen to know why they would ask for u and v in terms of x and y (instead of x and y in terms of u and v) and how this might be helpful in any way?
I don't know what their motivation was, for setting up the exercise as they did. Sometimes, exercises are designed solely for practice (eg: concepts, properties, symbolic reasoning). They're helpful for that, but such exercises don't always apply to real-world situations or serve as a discovery moment. Sometime, they don't even bridge to the next topic; you may need to wait for calculus, before using again some ideas learned in precalculus.

I would suggest asking your instructor these questions. 8-)
 
There's a sign error, in your expression for u. (When squaring x+i·y, don't forget that i gets squared, too.)

I had not combined the ratios, in my expression for u:

u = 1/(4y^4) - y^2

After correcting my careless error and manipulating the expressions I too had the same expression for u.
I have since tried further examples and your method works perfectly - these had been bugging me for a while but now I think I have a slightly deeper understanding of what's going on, so this was a big help!

Sometimes, exercises are designed solely for practice (eg: concepts, properties, symbolic reasoning).
I think you're right here. Especially since w is a function of z perhaps asking it seems somewhat more natural.

I would suggest asking your instructor these questions.
Unfortunately they aren't great :(

But nevertheless thanks for your advice! :D
 
Did you finish the second exercise? I suspect the locus of Q is an enlargement of the locus of P. :cool:
 
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