- Home
- Standard 11
- Mathematics
4-1.Complex numbers
normal
Let the minimum value $v_{0}$ of $v=|=|^{2}+|z-3|^{2}+|z-60|^{2}$, $z \in C$ is attained at $z=z_{0}$. Then $\left|2 z_{0}^{2}-z_{0}^{3}+3\right|^{2}+v_{0}^{2}$ is equal to.
A
$1000$
B
$1024$
C
$1105$
D
$1196$
(JEE MAIN-2022)
Solution
$z_{0} =\left(\frac{0+3+0}{3}, \frac{0+6+0}{3}\right)=(1,2)$
$v_{0}=|1+2 i|^{2}+|1+2 i-3|^{2}+|1+2 i-6 i|^{2}=30$
$\text { Then }\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$
$=\left|2(1+2 i)^{2}-(1-2 i)^{3}+3\right|^{2}+900$
$=|2(1-4+4 i)-(1-4-4 i)(1-2 i)+3|^{2}+900$
$=|8+6 i|^{2}+900=100+900=1000$
Standard 11
Mathematics
