Question

If z ≠ 0 be a complex number such that$|z-\frac{1}{z}|=2$, then the maximum value of |z| is

• A. $\sqrt2$
• B. 1
• C. $\sqrt2-1$
• D. $\sqrt2+1$

Answer 1

Answer: (D)

$\sqrt2+1$

Answer 2

$|z-\frac{1}{z}|$≥||z$|\frac{-1}{z}|$

⇒ $||z|-\frac{1}{|z|}|$≤2

Let |z | = r

$|r-\frac{1}{r}|$≤2

−2≤$r-\frac{1}{r}$≤2

$r-\frac{1}{r}$≥−2 and $r-\frac{1}{r}$≤2

$r^2$+2r–1≥0 and $r^2$–2r–1≤0

r∈[−∞,−1–$\sqrt2$]∪[−1+$\sqrt2$,∞] and r∈[1−$\sqrt2$, 1+$\sqrt2$]

Taking intersection r∈[$\sqrt2-1,\sqrt2+1$]
So, the correct option is (D): $\sqrt2+1$.