Question

The greatest and least value of $|z_1+z_2|$ if $z_1=24+7i$ and $\left|z_2\right|=6$ are respectively

• A. $31,25$
• B. $25,19$
• C. $31,19$
• D. None of these

Answer 1

Answer: (C)

$31,19$

$\left|z_2\right|=6$ represents a circle of radius $6$ centred at $(0,0)$ $z_1=24+7i$ represents a point $A(24,7)$.
Let a diameter through $A$meet the circle in points $B$ and $C$, then $AB$ is leastand $AC$ is greatest distance of $A$ from the circle. Clearly
$A{O}^2={24}^2+7^2=625$ $\therefore AO=25$
$\therefore Greatdistance=AC+AO+OC$
$=25+6=31$
$\therefore leastdistance=AB=AO-BO=25-6=19$
Alternative Solution.
Since,$\left|z^2\right|=6$ , $\therefore z_2$may be taken as $z_2=6e^{i\theta }$
or $z_2=6\left(\cos \theta +i\sin \theta \right),z_1=24+7i$
$\therefore z_1+z_2=(24+6\cos \theta )+i(7+6\sin \theta )$
$\Rightarrow [z_1+z_2{]}^2={(24+6\cos \theta )}^2+{(7+6\sin \theta )}^2$
$=576+49+12(24\cos \theta +7\sin \theta )$
$+36({\cos }^2\theta +{\sin }^2\theta )$
Now put $24=r\sin \alpha ,7=r\cos \alpha$
Where $r^2={24}^2+7^2=625$ $\therefore r=25$
$\therefore d^2=625+36+12r\sin (\theta +\alpha )$
$=661+12\left(25\right)(\pm 1):+formax.-formin$
$=661+300$ or $661-300$
$=961$ or $361$ i.e., $(31{)}^2$ or $(19{)}^2$
$\therefore d=31\left(max\right),=19(min.)$