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