Inequalities of complex numbers

Question

Solve the inequality 2 <= |z ^ 2 - 3| <= 4.

\(2 \le |z^2 -3| \le 4\)

Answer 1

\(2 \le |z^2 -3| \le 4\)

Let \(z = x + iy\)

\(z^2 = (x + iy)^2 = x^2 - y^2 + 2xyi\)

\(\therefore 2 \le |x^2 - y^2 + 2xyi - 3| \le 4\)

⇒ \(4 \le |x^2 - y^2 - 3 + 2xyi| ^2 \le 16\)

⇒ \(4 \le (x^2 - y^2 -3)^2 + 4x^2 y^2 \le 16\)

⇒ \(4 \le x^4 + y^4 + 9 - 2x^2 y^2 - 6x^2 + 6y^2 + 4x^2 y^2 \le 16\)

⇒ \(4 \le (x^2 + y^2)^2 - 6x^2 + 6y^2 + 9 \le 16\)

⇒ \(-5 \le (x^2 + y^2)^2 - 6x^2 + 6y^2 \le 7\)

Alternative:

Let \(w = z^2\in C\)

Then \(2 \le |w -3| \le 4\)

is a circular ring whose centre is (3, 0) and radius lies between (2, 4).