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+15i. Then |z_1^4+z_2^4| equals-

Given: {z}_{1}+{z}_{2}=5&{{z}_{1}}^{3}+{{z}_{2}}^{3}=20+15i

Now, solving ${{z}_{1}}^{3}+{{z}_{2}}^{3}=20+15i$z1​3+z2​3=20+15i

$$\Rightarrow {({z}_{1}+{z}_{2})}^{3}-3{z}_{1}{z}_{2}({z}_{1}+{z}_{2})=20+15i$$⇒(z1​+z2​)3−3z1​z2​(z1​+z2​)=20+15i

$$\Rightarrow {(5)}^{3}-3{z}_{1}{z}_{2}(5)=20+15i$$⇒(5)3−3z1​z2​(5)=20+15i

$$\Rightarrow 125-15{z}_{1}{z}_{2}=20+15i$$⇒125−15z1​z2​=20+15i

$$\Rightarrow {z}_{1}{z}_{2}=7-i$$⇒z1​z2​=7−i

Now, finding ${{z}_{1}}^{4}+{{z}_{2}}^{4}={({{z}_{1}}^{2}+{{z}_{2}}^{2})}^{2}-2{({z}_{1}{z}_{2})}^{2}$z1​4+z2​4=(z1​2+z2​2)2−2(z1​z2​)2

$$\Rightarrow {{z}_{1}}^{4}+{{z}_{2}}^{4}={({({z}_{1}+{z}_{2})}^{2}-2{z}_{1}{z}_{2})}^{2}-2{({z}_{1}{z}_{2})}^{2}$$⇒z1​4+z2​4=((z1​+z2​)2−2z1​z2​)2−2(z1​z2​)2

$$\Rightarrow {{z}_{1}}^{4}+{{z}_{2}}^{4}={(25-2(7-i))}^{2}-2{(7-i)}^{2}$$⇒z1​4+z2​4=(25−2(7−i))2−2(7−i)2

$$\Rightarrow {{z}_{1}}^{4}+{{z}_{2}}^{4}={(11+2i)}^{2}-2(48-14i)$$⇒z1​4+z2​4=(11+2i)2−2(48−14i)

$$\Rightarrow {{z}_{1}}^{4}+{{z}_{2}}^{4}=(121-4+44i)-96+28i$$⇒z1​4+z2​4=(121−4+44i)−96+28i

$$\Rightarrow {{z}_{1}}^{4}+{{z}_{2}}^{4}=21+72i$$⇒z1​4+z2​4=21+72i

$$\Rightarrow |{{z}_{1}}^{4}+{{z}_{2}}^{4}|=\sqrt{{21}^{2}+{72}^{2}}$$⇒∣z1​4+z2​4∣=212+722​

$$\Rightarrow |{{z}_{1}}^{4}+{{z}_{2}}^{4}|=\sqrt{5625}$$⇒∣z1​4+z2​4∣=5625​

$$\Rightarrow |{{z}_{1}}^{4}+{{z}_{2}}^{4}|=75$$⇒∣z1​4+z2​4∣=75