Let z _ 1 and z _ 2 be two complex number such that z _ 1 + z _ 2 =...

Given: ${z}_{1} + {z}_{2} = 5 \text{ and } {{z}_{1}}^{3} + {{z}_{2}}^{3} = 20 + 15i$

Now, solving ${{z}_{1}}^{3} + {{z}_{2}}^{3} = 20 + 15i$

$⇒ {\left({z}_{1} + {z}_{2}\right)}^{3} - 3{z}_{1}{z}_{2}\left({z}_{1} + {z}_{2}\right) = 20 + 15i$

$⇒ {\left(5\right)}^{3} - 3{z}_{1}{z}_{2}\left(5\right) = 20 + 15i$

$⇒ 125 - 15{z}_{1}{z}_{2} = 20 + 15i$

$⇒ {z}_{1}{z}_{2} = 7 - i$

Now, finding ${{z}_{1}}^{4} + {{z}_{2}}^{4} = {\left({{z}_{1}}^{2} + {{z}_{2}}^{2}\right)}^{2} - 2{\left({z}_{1}{z}_{2}\right)}^{2}$

$⇒ {{z}_{1}}^{4} + {{z}_{2}}^{4} = {\left({\left({z}_{1} + {z}_{2}\right)}^{2} - 2{z}_{1}{z}_{2}\right)}^{2} - 2{\left({z}_{1}{z}_{2}\right)}^{2}$

$⇒ {{z}_{1}}^{4} + {{z}_{2}}^{4} = {\left(25 - 2\left(7 - i\right)\right)}^{2} - 2{\left(7 - i\right)}^{2}$

$⇒ {{z}_{1}}^{4} + {{z}_{2}}^{4} = {\left(11 + 2i\right)}^{2} - 2\left(48 - 14i\right)$

$⇒ {{z}_{1}}^{4} + {{z}_{2}}^{4} = \left(121 - 4 + 44i\right) - 96 + 28i$

$⇒ {{z}_{1}}^{4} + {{z}_{2}}^{4} = 21 + 72i$

$⇒ \left|{{z}_{1}}^{4} + {{z}_{2}}^{4}\right| = \sqrt{{21}^{2} + {72}^{2}}$

$⇒ \left|{{z}_{1}}^{4} + {{z}_{2}}^{4}\right| = \sqrt{5625}$

$⇒ \left|{{z}_{1}}^{4} + {{z}_{2}}^{4}\right| = 75$