You visited us 1 times! Enjoying our articles? Unlock Full Access!

Logarithmic Inequalities

If |z2-3|=3...

Question

If $| z^{2} - 3 | = 3 | z |$ , then the maximum value of |z| is

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3 + \sqrt{21}}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{\sqrt{21} - 3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{3 + \sqrt{21}}{2}$
By the law of inequality,
$| z^{2} - 3 | \geq {| z |}^{2} - 3$
$⟹ 3 | z | \geq {| z |}^{2} - 3 ⟹ {| z |}^{2} - 3 | z | - 3 \leq 0 ⟹ 0 \leq | z | \leq \frac{3 + \sqrt{21}}{2}$
Hence the maximum value of $| z | = \frac{3 + \sqrt{21}}{2}$

Suggest Corrections

Similar questions

z = \frac{1 - i \sqrt{3}}{2}, i = \sqrt{- 1} .

21 + {(z + \frac{1}{z})}^{3} + {(z^{2} + \frac{1}{z^{2}})}^{3} + {(z^{3} + \frac{1}{z^{3}})}^{3} + \dots + {(z^{21} + \frac{1}{z^{21}})}^{3}

z^{2} + z + 1 = 0

(z + \frac{1}{z})^{2} + (z^{2} + \frac{1}{z^{2}})^{2} + (z^{3} + \frac{1}{z^{3}})^{2} + . . . . . . . . + (z^{21} + \frac{1}{z^{21}})^{2}

_{1}

=

_{2}

=

_{3}

=

_{1}

_{2}

_{1}

_{3}

_{2}

_{3}

=

| z_{1}

z_{2}

z_{3} |

z = \frac{- 2}{1 + i \sqrt{3}}

\frac{π}{3}

\frac{2 π}{3}

\frac{π}{4}

| z |

z

| z + \frac{2}{z} | = 2

1 + \sqrt{3}

| z_{1} + z_{2} | \leq | z_{1} | + | z_{2} |

Join BYJU'S Learning Program