Latest revision as of 16:37, 4 October 2020

Let $\mathcal{T}$ be the locus of all points $z$ in the complex plane satisfying $|\text{Re}(z)|=\text{Im}(z)+1$ , and let $\mathcal{T}'$ be the locus of all points $z'=\frac{i}{z}$ where $z \in \mathcal{T}$ and $i=\sqrt{-1}$ . If the area enclosed by $\mathcal{T}'$ is $\mathcal{A}$ , compute $\lfloor 100\mathcal{A} \rfloor$ .

Solution 1

Graph the equation $|\text{Re}(z)|=\text{Im}(z)+1$ in the complex plane and you will find that the locus of all points $\frac{i}{z}$ is the intersection of two circles and has area $75\pi+50$ . The greatest integer less than or equal to $75\pi+50$ is $\boxed{285}$ .

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 =Solution 1=
-Graph the equation <math>|\text{Re}(z)|=\text{Im}(z)+1</math> in the complex plane and you will find that the locus of all points <math>\frac{i}{z}</math> is the intersection of two circles and has area <math>75π+50</math>. The greatest integer less than or equal to <math>75π+50</math> is <math>\boxed{285}</math>.
+Graph the equation <math>|\text{Re}(z)|=\text{Im}(z)+1</math> in the complex plane and you will find that the locus of all points <math>\frac{i}{z}</math> is the intersection of two circles and has area <math>75\pi+50</math>. The greatest integer less than or equal to <math>75\pi+50</math> is <math>\boxed{285}</math>.
 ==See also==

2019 CIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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Difference between revisions of "2019 CIME I Problems/Problem 12"

Latest revision as of 16:37, 4 October 2020

Solution 1

See also