If
is a purely imaginary number (z ≠ –1), then find the value of |z|.
Let z = x + iy
Now, ![]()
![]()
![]()
Given that
is purely imaginary.
![]()
⇒ x2 – 1 + y2 = 0
⇒ x2 + y2 = 1
![]()
∴ |z| = 1
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