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

 Jun 7, 2021
 #1
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z = a + bi

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

sqrt(a^2 - 6a + 9 + b^2) = sqrt(a^2 + b^2 + 2b + 1)

a^2 - 6a + 9 + b^2 = a^2 + b^2 + 2b + 1

-6a + 9 = 2b + 1

8 = 6a + 2b + 1

7 = 6a + 2b

 

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

|a + (b+1)i| = |(a-1) + (b-2)i|

sqrt(a^2 + b^2 + 2b + 1)  = sqrt(a^2 - 2a + 1 + b^2 - 4b + 4)

a^2 + b^2 + 2b + 1 = a^2 - 2a + 1 + b^2 - 4b + 4

2b + 1 = - 2a + 1 - 4b + 4

6b + 2a = 4

3b + a = 4

 

7 = 6a + 2b

3b + a = 4

a = 4 - 3b

7 = 6(4 - 3b) + 2b

7 = 24 - 18b + 2b

16b = 24 - 18b + 2b

16b = 17

b = 17/16

a = 13/16

 

There are probably other ones, but 13/16 + 17/16i is the one I found. 

 

=^._.^=

 Jun 7, 2021

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