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# Find all complex numbers Z such that... PLZ HELP ME!!!

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Find all complex numbers Z such that $$|z - 3| = |z+i| = |z -3i|.$$

Oct 12, 2019

#1
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We are looking for all complex numbers that are equidistant from  3, -i and 3i. Since these three points are not collinear, we know there is a unique such point (the circumcenter of the triangle formed by the three numbers in the complex plane).

The numbers equidistant from  and  are on the perpendicular bisector of the segment connecting -i and 3i in the plane. This perpendicular bisector is Im(z)=1, so the imaginary part of  is 1.

The numbers equidistant from 3 and 3i are on the perpendicular bisector of the segment connecting 3 and 3i in the plane. This perpendicular bisector is simply Re(z)=Im(z). Combining this with Im(z)=1 tells us that z=1+i.

Oct 12, 2019
#2
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Find all complex numbers Z such that    |z - 3| = |z+i| = |z -3i|

Let z=a+bi

So I want

$$|a+bi - 3| = |a+bi+i| = |a+bi -3i|\\ |(a-3)+bi | = |a+(b+1)i| = |a+(b-3)i |\\ (a-3)^2+b^2=a^2+(b+1)^2=a^2+(b-3)^2\\ a^2-6a+9+b^2=a^2+b^2+2b+1=a^2+b^2-6b+9\\ -6a+9=2b+1=-6b+9\\~\\ 2b+1=-6b+9\\ 8b=8\\ b=1\\~\\ -6a+9=2b+1\\ -6a+9=2*1+1\\ -6a+9=3\\ -6a=-6\\ a=1$$

So the only z I get is      $$\bf {z=1+i}$$

check

$$|z - 3| = |z+i| = |z -3i|\\ |1+i - 3| = |1+i+i| = |1+i -3i|\\ |-2+i | = |1+2i| = |1-2i|\\ \sqrt5 = \sqrt5 =\sqrt5$$

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Oct 12, 2019
#3
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....

Oct 12, 2019
edited by Rom  Oct 18, 2019
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Hi Rom,

It is always frustrating when this happens but I think these 2 questions have been asked by different people.

It was Oofrence who asked last time, and you that answered. Maybe he/she learned from you.

Oct 12, 2019
#5
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....

Rom  Oct 12, 2019
edited by Rom  Oct 18, 2019
#6
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They are probably kids in the same class.

It would be good if  we could do something like that but this site is not organised enough to manage it.

The kids would not look anyway. Melody  Oct 13, 2019