JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q1

MathematicsSingle correct+4 / -1Question 1 of 90

If z ≠ 0 be a complex number such that $\left|z-\dfrac1z\right|=2$ , then the maximum value of |z| is :

\sqrt2

\sqrt2-1

\sqrt2+1

Answer and worked solution

Official answer: D

Triangle inequality. $|z|-\frac1{|z|}\le\left|z-\frac1z\right|=2$ , so $|z|^2-2|z|-1\le0$ and $|z|\le1+\sqrt2$ (attained at $z=(1+\sqrt2)i$ ).

(D)

Topics:Modulus and argument

JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 1