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Find a function f(z) such that the line below is the graph of the equation |z-3i|=|f(z)|. z is a complex number in the form a+bi in all 3 questions

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i know the equation of the line is y = x/2 + 1/2, but i don't know how to continue on

 

 

The square of the non-real complex number \(z\) is equal to \(\overline{z}\). What is the real part of z?

i wrote an equation that was \((a+bi)^2=a-bi\) and from there \(a^2 + 2abi - b^2 = a - bi\), but i don't know how to continue on from there

 

 

Find all complex numbers z such that\(|z - 3| = |z+i| = |z -3i|.\)

i tried writing some equations but it didn't work out

 Oct 8, 2019
 #1
avatar+974 
+1

Sorry ... don't know how to do this!

 Oct 8, 2019
 #2
avatar+6045 
+3

#2

 

\(z^2 = \bar{z}\\ \text{it should be intuitively clear that $|z| = 1$ so let $z = e^{i\theta}$}\\ e^{i2\theta} = e^{-i\theta + 2\pi k},~k \in \mathbb{Z}\\ 2\theta = 2\pi - \theta\\ \theta = \dfrac{2\pi}{3}\\ z = e^{i2\pi/3} = -\dfrac 1 2 + i \dfrac{\sqrt{3}}{2}\)

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 Oct 8, 2019
 #3
avatar+6045 
+3

#3

 

\(|z-3|=|z+i|=|z-3i|\\ \text{let $z=x+i y$}\\ (x-3)^2 +y^2 = x^2 + (y-3)^2\\ x^2 - 6x+9 +y^2 = x^2 + y^2 - 6y+9\\ -6x=-6y\\ x = y\)

 

\(|x-3+xi| = |x + (1+x)i|\\ (x-3)^2 + x^2 = x^2 + (x+1)^2\\ x^2-6x+9 = x^2 + 2x + 1\\ 8=8x\\ x=1\\ z = 1+i\)

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 Oct 8, 2019

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