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# Complex Numbers

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Find all complex numbers z such that |z - 3| = |z + i| = |z - 1 - 2i|.

Apr 8, 2022

#1
0

z = 4i

sub in

|4i - 3| = |4i + i| = |4i - 1 - 2i| = |4i -  3| = |5i| = |2i - 1| so

z = i and 5

Apr 8, 2022
#2
+117180
+1

|z - 3| = |z + i| = |z - 1 - 2i|.

let z= a+bi

$$|a+bi - 3| = |a+bi + i| = |a+bi - 1 - 2i| \\ |(a-3)+bi| = |a+(b+1)i| = | (a -1)+(b - 2)i |\\ (a-3)^2+b^2 = a^2+(b+1)^2 = (a-1)^2+(b-2)^2\\ a^2-6a+9+b^2=a^2+b^2+2b+1=a^2-2a+1+b^2-4b+4\\ -6a+9=2b+1=-2a+1-4b+4\\ -6a+8=2b=-2a-4b+4\\ -3a+4=b=-a-2b+2\\~\\ b=-3a+4\\so\\ -3a+4=-a-2(-3a+4)+2\\ -3a+4=-a+6a-8+2\\ -3a+4=5a-6\\ -8a=-10\\ a=1.25\\ b=-3a+4=-3.75+4=0.25\\ z=\frac{5+i}{4}$$

Latex:

|a+bi - 3| = |a+bi + i| = |a+bi - 1 - 2i| \\
|(a-3)+bi| = |a+(b+1)i| = | (a -1)+(b - 2)i |\\

(a-3)^2+b^2 = a^2+(b+1)^2 = (a-1)^2+(b-2)^2\\
a^2-6a+9+b^2=a^2+b^2+2b+1=a^2-2a+1+b^2-4b+4\\
-6a+9=2b+1=-2a+1-4b+4\\
-6a+8=2b=-2a-4b+4\\
-3a+4=b=-a-2b+2\\~\\
b=-3a+4\\so\\
-3a+4=-a-2(-3a+4)+2\\
-3a+4=-a+6a-8+2\\
-3a+4=5a-6\\
-8a=-10\\
a=1.25\\
b=-3a+4=-3.75+4=0.25\\
z=\frac{5+i}{4}

Apr 8, 2022
edited by Melody  Apr 8, 2022