Let z_1 and z_2 be two complex numbers such that z_1+z_2=5 and z_3^1+z_3^2=20+15i Then, |z…

Question

Let ${z}_{1}$ and ${z}_{2}$ be two complex numbers such that ${z}_{1}+{z}_{2}=5$ and ${z}_{3}^{1}+{z}_{3}^{2}=20+15i$ Then, $|{z}_{4}^{1}+{z}_{4}^{2}|$ equals -

Accepted Answer

$\begin{gathered} {z}_{1}+{z}_{2}=5 \ {z}_{3}^{1}+{z}_{3}^{2}=20+15i \ {z}_{3}^{1}+{z}_{3}^{2}={({z}_{1}+{z}_{2})}^{3}-3{z}_{1}{z}_{2}({z}_{1}+{z}_{2}) \ {z}_{3}^{1}+{z}_{3}^{2}=125-3{z}_{1}\cdot {z}_{2}(5) \ \Rightarrow 20+15i=125-15{z}_{1}{z}_{2} \ \Rightarrow 3{z}_{1}{z}_{2}=25-4-3i \ \Rightarrow 3{z}_{1}{z}_{2}=21-3i \ \Rightarrow {z}_{1}\cdot {z}_{2}=7-i \ \Rightarrow {({z}_{1}+{z}_{2})}^{2}=25 \ \Rightarrow {z}_{2}^{1}+{z}_{2}^{2}=25-2(7-i) \ \Rightarrow 11+2i \ {({z}_{2}^{1}+{z}_{2}^{2})}^{2}=121-4+44i \ \Rightarrow {z}_{4}^{1}+{z}_{4}^{2}+2(7-i{)}^{2}=117+44i \ \Rightarrow {z}_{4}^{1}+{z}_{4}^{2}=117+44i-2(49-1-14i) \ \Rightarrow |{z}_{4}^{1}+{z}_{4}^{2}|=75 \end{gathered}$

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