Let z_1 and z_2 be two complex numbers such t | vedclass

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(B) Given $z_1 + z_2 = 5$ and $z_1^3 + z_2^3 = 20 + 15i$.Using the identity $z_1^3 + z_2^3 = (z_1 + z_2)^3 - 3z_1z_2(z_1 + z_2)$:$20 + 15i = (5)^3 - 3

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$75$

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(B) Given $z_1 + z_2 = 5$ and $z_1^3 + z_2^3 = 20 + 15i$.Using the identity $z_1^3 + z_2^3 = (z_1 + z_2)^3 - 3z_1z_2(z_1 + z_2)$:$20 + 15i = (5)^3 - 3z_1z_2(5)$$20 + 15i = 125 - 15z_1z_2$$15z_1z_2 = 105 - 15i$$z_1z_2 = 7 - i$Now,$z_1^2 + z_2^2 = (z_1 + z_2)^2 - 2z_1z_2 = 25 - 2(7 - i) = 25 - 14 + 2i = 11 + 2i$.Squaring both sides:$(z_1^2 + z_2^2)^2 = (11 + 2i)^2 = 121 + 44i - 4 = 117 + 44i$.Also,$(z_1^2 + z_2^2)^2 = z_1^4 + z_2^4 + 2(z_1z_2)^2$.$z_1^4 + z_2^4 = 117 + 44i - 2(7 - i)^2$$z_1^4 + z_2^4 = 117 + 44i - 2(49 - 1 - 14i) = 117 + 44i - 2(48 - 14i)$$z_1^4 + z_2^4 = 117 + 44i - 96 + 28i = 21 + 72i$.Wait,let us re-evaluate: $z_1^4 + z_2^4 = (z_1^2 + z_2^2)^2 - 2(z_1z_2)^2 = (11 + 2i)^2 - 2(7 - i)^2 = (121 - 4 + 44i) - 2(49 - 1 - 14i) = 117 + 44i - 2(48 - 14i) = 117 + 44i - 96 + 28i = 21 + 72i$.Re-calculating the modulus: $|21 + 72i| = \sqrt{21^2 + 72^2} = \sqrt{441 + 5184} = \sqrt{5625} = 75$.

Let $z_1$ and $z_2$ be two complex numbers such that $z_1 + z_2 = 5$ and $z_1^3 + z_2^3 = 20 + 15i$. Then $|z_1^4 + z_2^4|$ equals-

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