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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 \(b\)

 Dec 20, 2018
 #1
avatar+3563 
+1

\(a=\dfrac 1 2\\ b = \dfrac{\sqrt{35}}{2}\)

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 Dec 20, 2018
 #2
avatar+474 
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can i see the steps please?

RektTheNoob  Dec 21, 2018
 #3
avatar+3563 
+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

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