# 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 $$f(\Omega ) = \{ f(z) : z \in \Omega \}$$

##### Source of the problem

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

##### Topic

Complex Numbers, Level Curves

5 out 10

##### Suggested Book

Complex Numbers from A to Z by Titu Andreescu

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. 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.

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

Since

$$u = x^2 – y^2 + 2$$

$$v = 2xy$$

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. # Connected Program at Cheenta

#### I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are: B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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