A function $f$ is defined on the complex numbers by $f(z) = (a + bi)z,$ where $a$ and $b$ are positive real numbers. This function has the

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A function $f$ is defined on the complex numbers by $f(z) = (a + bi)z,$ where $a$ and $b$ are positive real numbers. This function has the

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A function $f$ is defined on the complex numbers by

$f(z) = (a + bi)z,$

where $a$ and $b$ are positive real numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that

$|a + bi| = 3$

find $a$ and $b$

RektTheNoob Dec 20, 2018

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Rom · Answer 1 · 2018-12-20T08:40:55

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$a=\dfrac 1 2\\ b = \dfrac{\sqrt{35}}{2}$

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Rom Dec 20, 2018

#2

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can i see the steps please?

RektTheNoob Dec 21, 2018

#3

+6251

+1

I'll give you a hint.

$\text{since }f(z) \text{ has the stated property }\forall z\\ f(1) \text{ is equidistant from }(0,0i), \text{ and }(1,0i)\\ \text{The locus of these points is }\left(\dfrac 1 2 , i y\right),~\forall y \in \mathbb{R}\\ \text{From this you can figure out }a\\ \text{Once you have }a \text{ finding }b \text{ is trivial}$

Rom Dec 21, 2018

edited by Rom Dec 21, 2018

A function $f$ is defined on the complex numbers by $f(z) = (a + bi)z,$ where $a$ and $b$ are positive real numbers. This function has the

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A function ff is defined on the complex numbers by f(z)=(a+bi)z,f(z) = (a + bi)z,where aa and bb are positive real numbers. This function has the

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A function $f$ is defined on the complex numbers by $f(z) = (a + bi)z,$ where $a$ and $b$ are positive real numbers. This function has the

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