+0  
 
-1
43
1
avatar+7 

Hi! Can someone please solve the below problem?

 

 

 

 

 

 

 

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

 

Express each answer in the form $a+bi$, where $a$ and $b$ are real numbers.

 Jan 13, 2021
 #1
avatar+112034 
+2

Finish solving this.

I only got one answer but maybe there are more.

 

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

 Jan 13, 2021

15 Online Users