ISI 2019 : Problem #3

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ISI 2019 : Problem #3

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Indian Statistical Institute

complex numbers

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Source: I.S.I. 2019

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3465 posts

#1 h May 5, 2019, 11:52 AM • 3 Y

Y by Adventure10, Mango247, PikaPika999

Let $\Omega=\{z=x+iy~\in\mathbb{C}~:~|y|\leqslant 1\}$ . If $f(z)=z^2+2$ , then draw a sketch of $f\Big(\Omega\Big)=\{f(z):z\in\Omega\}$ Justify your answer.

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#2 h May 5, 2019, 12:20 PM • 3 Y

Y by Adventure10, Mango247, PikaPika999

Is it a disc with centre at (2,0) and radius 1 ?

EDIT : Sorry, read the question wrong, (Assumed locus of z is a disc with rad 1 and centre at O)

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#3 h May 5, 2019, 12:47 PM • 3 Y

Y by Adventure10, Mango247, PikaPika999

It's not a disk because it is not bounded. I'll do $g(z)=z^2$ . We have $g(x+i)=(x+i)^2=x^2-1+2ix$ which is the curve $y^2-4x=4$ . Also $g(-x-i)$ has the same image since $g$ is even. So $g(\Omega)$ is the set bounded by $y^2-4x=4$ (a parabola) which contains $(0,0)$ .

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#4 h May 5, 2019, 1:53 PM • 2 Y

Y by math_and_me, Adventure10

It was g(z)=z^2+2 ...hence i think the parabola should be y^2=4x-4

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#5 h May 5, 2019, 1:59 PM • 4 Y

Y by danepale, Adventure10, Mango247, PikaPika999

No, I took $g(z)=z^2$ , that's why I used a different letter. If you consider $f$ then the bounding parabola would be $(y-2)^2-4x=4$ since we do a translation $T{0\choose 2}$ .

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#6 h May 5, 2019, 2:38 PM • 5 Y

Y by Adventure10, Mango247, Mango247, Mango247, PikaPika999

Not qualifying ISI entrance for sure! Done only 3 subjective correctly any 23 mcq. So goodbye aops.

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SRHMEC

#8 h Jan 16, 2020, 8:55 PM • 3 Y

Y by Adventure10, Mango247, PikaPika999

I couldn't understand from where to start and so leave about the end. How to plot this?

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majju

#9 h May 2, 2020, 9:17 PM • 2 Y

Y by Mango247, PikaPika999

The function is a complex function that takes complex inputs and gives complex outputs so shouldn't the graph be a 4-d one ?

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#10 h May 2, 2020, 10:16 PM

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I doesn't ask for the graph but for the image set. That is just a subset of ${\bf C}$ .

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#11 h May 3, 2020, 10:51 AM • 3 Y

Y by ankitgayen2018, AK001, A.T.G

$\Omega = \{z= x+iy \in \mathbb{C} : | y | \leq 1\}$ $f(z)=z^{2} +2$ $\text{Taking our form of complex numbers in}$ $\Omega$ $\text{we get that}$
$f(z)= x^{2}-y^{2}+2 +2xyi$ $\text{So}$ $Re(f(z))=x^{2}-y^{2} +2$ and $Im(f(z))= 2xy$ $\text{f(z) is nothing but a transformation T in}$ $\mathbb{C}$
$\text{T is such that}$ $(x,y) \mapsto (x^{2} -y^{2}+2,2xy)$ $Fix~ y \in \mathbb{R}$ $\text{Let us deote the complex plane as the u-v plane u denotes the real axis and v the imaginary axis}$ $\text{so we have the equations}$
$u=x^{2}-y^{2}+2$
$v=2xy$
$Hence~ \frac{u+y^{2}-2}{v^2}=\frac{1}{4y^{2}}$ $\text{So}$ $4y^{2}u +4y^{4}-8y^{2}=v^2$ $\text{It is given}$ $|y|\leq 1$ So it is the family of parabolas in $\mathbb{C}$ satisfying this condition $\text{In other words it is the family of parabolas such that their verex is at}$ $(2-y^{2},0)$
Look at this animation for the full family intersection of blue and gren gives us solution https://www.geogebra.org/graphing/bbfptpnb

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#12 h May 3, 2020, 11:19 AM • 1 Y

Y by A.T.G

Here is some visual help

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Raunii

#13 h Jul 14, 2020, 12:59 AM

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Mathcollege wrote:

Not qualifying ISI entrance for sure! Done only 3 subjective correctly any 23 mcq. So goodbye aops.

How was the result

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337 posts

#14 h Sep 17, 2020, 3:55 PM

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Agsh2005 wrote:

$\Omega = \{z= x+iy \in \mathbb{C} : | y | \leq 1\}$ $f(z)=z^{2} +2$ $\text{Taking our form of complex numbers in}$ $\Omega$ $\text{we get that}$
$f(z)= x^{2}-y^{2}+2 +2xyi$ $\text{So}$ $Re(f(z))=x^{2}-y^{2} +2$ and $Im(f(z))= 2xy$ $\text{f(z) is nothing but a transformation T in}$ $\mathbb{C}$
$\text{T is such that}$ $(x,y) \mapsto (x^{2} -y^{2}+2,2xy)$ $Fix~ y \in \mathbb{R}$ $\text{Let us deote the complex plane as the u-v plane u denotes the real axis and v the imaginary axis}$ $\text{so we have the equations}$
$u=x^{2}-y^{2}+2$
$v=2xy$
$Hence~ \frac{u+y^{2}-2}{v^2}=\frac{1}{4y^{2}}$ $\text{So}$ $4y^{2}u +4y^{4}-8y^{2}=v^2$ $\text{It is given}$ $|y|\leq 1$ So it is the family of parabolas in $\mathbb{C}$ satisfying this condition $\text{In other words it is the family of parabolas such that their verex is at}$ $(2-y^{2},0)$
Look at this animation for the full family intersection of blue and gren gives us solution https://www.geogebra.org/graphing/bbfptpnb

Isn't the $x$ coordinate of the vertex of a parabola $-\frac{b}{2a}$ ? In your case, $b=0$ .

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lifeismathematics

1188 posts

lifeismathematics

#16 h Jul 6, 2023, 1:36 PM • 1 Y

Y by kamatadu

we have $f(z)=x^2-y^2+2+2ixy$ so this is a transformation: $(x,y) \mapsto (x^2-y^2+2,2xy)$

now we try to find image of $f(z)$ which is $f(\Omega)$ under complex plane, so let's denote the complex plane as $\alpha-\beta$ plane where $\alpha=x^2-y^2+2, \beta=2xy$ , now we have $\alpha=\left(\frac{\beta}{2y}\right)^2+2-y^2$

so $f(\Omega)$ denote a family of parabola(s) having common vertex at $(2-y^2,0)$ for $|y| \leqslant 1.$

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#17 h Apr 6, 2025, 2:24 PM

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To me, it was very confusing at first why we are defining a new $u$ - $v$ plane, but I'll try to explain it here.

Since $f(z)$ takes input in $\mathbb{C}$ and outputs in $\mathbb{C}$ , we can't just draw a typical $\mathbb{R} \to \mathbb{R}$ graph and change the $y$ -axis's name.

So, $f(z)$ is transforming the entire plane of complex numbers, and that transformation is given by:
$(x, y) \mapsto (x^2 - y^2 + 2,\ 2xy)$
Basically, it can take input from anywhere in the Argand plane, not just from one of the axes. So we can say that the function transforms the entire Argand plane.

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#18 h Apr 6, 2025, 3:36 PM • 1 Y

Y by PikaPika999

sketch of the graph

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