The least value of |z| where z is complex number which satisfies the inequality exp ( (|z| + 3)(|z| - 1)/(||z| + 1|)log ₑ2 ) ≥ log _(√2)|5√7 + 9i|,i = √(- 1) , is equal to :

Solution

Maths sir · 4 steps

Turn both sides into powers of 2. The inequality then becomes a simple quadratic in |z|.

1. Right side. |5√7 + 9i| = √(175 + 81) = √256 = 16. Since 16 = 2⁴ = (√2)⁸, we get log_(√2) 16 = 8 = 2³.
2. Left side. Use eˣˡⁿ² = 2ˣ. Put r = |z| ≥ 0. Then |r + 1| = r + 1, and the left side is 2^((r+3)(r-1)/(r+1)).
3. Compare exponents. The base 2 is greater than 1, so we can compare exponents directly. Since r + 1 > 0, multiply through by it without flipping the sign.

   (r+3)(r-1)/(r+1) ≥ 3 ⇒ r² + 2r - 3 ≥ 3r + 3 ⇒ (r-3)(r+2) ≥ 0

4. So. With r ≥ 0, the factor r + 2 is positive, so r ≥ 3. The least value of |z| is 3, option (b).

The idea to keep. When an exponential sits on one side and a logarithm on the other, rewrite both with the same base. Then compare exponents, and remember that |z| ≥ 0 fixes the sign of every factor.

Answer from the NTA final answer key. Solution by PrepWiser.