Maths · Modulus and argument (or amplitude) of a complex number
If then lies on
If \( \left|\mathbf{z}^{2}-\mathbf{1}\right|=|\mathbf{z}|^{2}+\mathbf{1}, \) then \( \mathbf{z} \) lies on
- A. the real axis
- B. the imaginary axis
- C. a circle
- D. an ellipse
Step-by-step solution
Let z = x + iy. Then |z|^2 = x^2 + y^2 and |z^2 - 1| = sqrt((x^2 - y^2 - 1)^2 + (2xy)^2) = sqrt((x^2 + y^2)^2 - 2x^2 + 2y^2 + 1). The equation becomes sqrt((x^2 + y^2)^2 - 2x^2 + 2y^2 + 1) = x^2 + y^2 + 1. Squaring both sides and simplifying yields -4x^2 = 0, so x = 0. Hence z lies on the imaginary axis.