Complex Number Question 14
Question 14 - 2024 (31 Jan Shift 2)
Let $z_{1}$ and $z_{2}$ be two complex number such that $z_{1}+z_{2}=5$ and $z_{1}^{3}+z_{2}^{3}=20+15 i$. Then $\left|z_{1}^{4}+z_{2}^{4}\right|$ equals-
(1) $30 \sqrt{3}$
(2) 75
(3) $15 \sqrt{15}$
(4) $25 \sqrt{3}$
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Answer (2)
Solution
$$ \begin{aligned} & z_{1}+z_{2}=5 \ & z_{1}^{3}+z_{2}^{3}=20+15 i \ & z_{1}^{3}+z_{2}^{3}=\left(z_{1}+z_{2}\right)^{3}-3 z_{1} z_{2}\left(z_{1}+z_{2}\right) \ & z_{1}^{3}+z_{2}^{3}=125-3 z_{1} \cdot z_{2}(5) \ & \Rightarrow 20+15 i=125-15 z_{1} z_{2} \ & \Rightarrow 3 z_{1} z_{2}=25-4-3 i \ & \Rightarrow 3 z_{1} z_{2}=21-3 i \ & \Rightarrow z_{1} \cdot z_{2}=7-i \ & \Rightarrow\left(z_{1}+z_{2}\right)^{2}=25 \ & \Rightarrow z_{1}^{2}+z_{2}^{2}=25-2(7-i) \ & \Rightarrow 11+2 i \ & \left(z_{1}^{2}+z_{2}^{2}\right)^{2}=121-4+44 i \ & \Rightarrow z_{1}^{4}+z_{2}^{4}+2(7-i)^{2}=117+44 i \ & \Rightarrow z_{1}^{4}+z_{2}^{4}=117+44 i-2(49-1-14 i) \ & \Rightarrow\left|z_{1}^{4}+z_{2}^{4}\right|=75 \end{aligned} $$
