Question

If a is a complex no. such that, |z| greater than or equal to 2, then the minimum value of |z+1/2| is
a. Strictly greatest than 5/2
b. Greater than 3/2 but less than 5/2
c. Equal to 5/2
d. Lies in the interval (1,2)

Answer 1

Step-by-step explanation:

$z \geqslant 2 \\ (z + \frac{1}{2} ) \\ (\geqslant 2 + \frac{1}{2} ) \\ if \: you \: observe \: it \: correctly \: we \: can \: easily \: conclude => \\ \: it \: is \: \geqslant \: \frac{5}{2} \\ so \: option \: (a) \: is \: correct$

Answer 2

Answer:

Option d

Let

$z = a \cos( \alpha ) +a i \sin( \alpha ) ) \\ |z| = a \: \geqslant 2 \\ z + \frac{1}{2} =( a \cos( \alpha ) + \frac{1}{2} )+a i \sin( \alpha ) ) \\ |z + \frac{1}{2} | = \sqrt{(a \cos( \alpha ) + \frac{1}{2}) {}^{2} + {(a \sin( \alpha )) }^{2} } \\ |z + \frac{1}{2} | = \sqrt{a {}^{2} \cos {}^{2} ( \alpha ) + \frac{1}{4} + {(a \sin( \alpha )) }^{2} + a \cos( \alpha ) } \\ |z + \frac{1}{2} | = \sqrt{ {a}^{2} + \frac{1}{4} + a \cos( \alpha ) }$

For minimum value alpha should be

90 or cos 90 = 0

**(You could also put this value in start to reduce calculations)

So

$min |z + \frac{1}{2} | = \sqrt{ {a}^{2} + \frac{1}{4} } \\ since \: \: \: \: \: a \geqslant 2 \\ min |z + \frac{1}{2} | \geqslant \sqrt{ {2}^{2} + \frac{1}{4} } = \frac{3}{2}$