Question

The number of complex numbers z satisfying |z-2-i| =|z-8+i| and |z+3|=1 is​

Answer 1

Answer:

z+∣z∣=2+8i⟶(1)

let z=x+iy⟶(2)

then,  ∣z∣=  

x  

2

+y  

2

 

​

⟶(3)

∴   Substituting (2) & (3) in (1) we get

x+iy+  

x  

2

+y  

2

 

​

=2+8i

⇒x+  

x  

2

+y  

2

 

​

+iy=2+8i

Comparing LHS & RHS we get

y=8⟶(4) and x+  

x  

2

+y  

2

 

​

=2⟶(5)

Substituting the value of y in the equation (5)

We get

x+  

x  

2

+64

​

=2

x  

2

+64

​

=2−x

Squaring both sides,

x  

2

+64=4+x  

2

−4x

⇒4x=−60

⇒x=−15

∴   z=−15+8i

∣z∣=  

225+164

​

=  

289

​

=17 units.