Question

The complex number z having least positive argument which satisfies the condition $|z-25i|\le 15$ is:

• A. $25i$
• B. $12+5i$
• C. $16+12i$
• D. $12+16i$

Answer 1

The correct option is D $12+16i$

Solution:

$|z-25i|\le 15$

Let $z=r(cos\theta +isin\theta )$

$\theta$ must be minimum

$|rcos\theta +i(rsin\theta -25)|\le 15$

$|\sqrt{r^{2}co{s}^2\theta +r^{2}si{n}^2\theta +625-50rsin\theta }|\le 15$

square both side

$r^2(co{s}^2\theta +si{n}^2\theta )+625-50rsin\theta \le 225$

$r^{2}50rsin\theta \le -400$

$f\left(r\right)=\frac{400+r^2}{50r}\le sin\theta$

Find maximum value of $f\left(r\right)=\frac{400+r^2}{50r}$

$f^{\prime }\left(r\right)=\frac{100r^2-50(400+r^2)}{2500r^2}=0$

$50r^2-50\times 400=0$

$r=20$

$f(r=20)=\frac{800}{1000}\le sin\theta$

$\frac{4}{5}\le sin\theta$

Least value of $sin\theta$ is $4/5$

$tan\theta =4/3cos\theta =3/5$

$z=20(3/5+4/5i)$

$z=12+16i$

D is correct.