Understand the problem

Let \( \Omega = \{ z = x + i y \ \mathbb{C} : |y| \leq 1 \} \). If \( f(z) = z^2 + 2 \) then draw a sketch of $$ f(\Omega ) = \{ f(z) : z \in \Omega \} $$

Justify your answer.

Source of the problem

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 3 from 2019 


Complex Numbers, Level Curves

Difficulty Level

5 out 10

Suggested Book

Complex Numbers from A to Z by Titu Andreescu

Start with hints

Do you really need a hint? Try it first!

Understand that the map is from x-y plane to u-v plane. (That is the input of the function is from a two-dimensional place and out put is also in ‘another’ two-dimensional plane.

The domain includes all points (x, y) such that \( -1 \leq y \leq 1 \). Here is a picture of the domain.

Domain of f(z)

We know that $$ f(z) = z^2 + 2 $$

Start with z = x + iy.

Then \( f(z) = (x + i y)^2 + 2 = x^2 – y^2 + 2 + 2xy i \) 

Hence in the output space: 

\( u = x^2 – y^2 + 2 \)

\( v = 2xy \) 

Draw level sets. That is, first find what happens at y = 0.

We know that

\( u = x^2 – y^2 + 2 \)

\( v = 2xy \)

Set y = 0 to get 

\( u = x^2 + 2 \)

\( v = 0 \)

Hence this is simply the horizontal ray starting at (2, 0) in the u-v plane.

Level set at y = 0

Level set at y = 0

Finally, find the different level curves by setting y = constant (and then varying this constant between -1 and 1)


\( u = x^2 – y^2 + 2 \)

\( v = 2xy \)

Hence \( \frac {v}{2y} = x \)

Replacing in first equation we have 

\( u = \frac{v^2}{4y^2} – y^2 + 2 \) 

(Notice y is not 0 as we have handled that case previously). 

This is a (family of) parabola(s) in the u-v plane with vertex at \( (2 – y^2, 0) \) and opening to the right. 

isi entrance 2019 subjective problem 3 solution


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