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Complex Numbers question

2024 · 31 Jan · Shift 2 · Q45
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  5. /2024 · 31 Jan · Shift 2 · Q45

Complex Numbers question

2024 · 31 Jan · Shift 2 · Q45

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let z1z_1z1​ and z2z_2z2​ be two complex numbers such that z1+z2=5z_1+z_2=5z1​+z2​=5 and z13+z23=20+15iz_1^3+z_2^3=20+15 iz13​+z23​=20+15i Then, ∣z14+z24∣\left|z_1^4+z_2^4\right|​z14​+z24​​ equals -
  1. A
    151515 \sqrt{15}1515​
  2. B
    30330 \sqrt{3}303​
  3. C
    25325 \sqrt{3}253​
  4. D
    75
View written solutionFree

Correct answer: D

  1. Let

    \qquad p=z_1z_2.$$

  2. Use the identity z13+z23=(z1+z2)3−3z1z2(z1+z2).z_1^3+z_2^3=(z_1+z_2)^3-3z_1z_2(z_1+z_2).z13​+z23​=(z1​+z2​)3−3z1​z2​(z1​+z2​). Substituting the given values, 20+15i=53−3p⋅5=125−15p.20+15i=5^3-3p\cdot 5=125-15p.20+15i=53−3p⋅5=125−15p. Hence, 15p=125−(20+15i)=105−15i,15p=125-(20+15i)=105-15i,15p=125−(20+15i)=105−15i, so p=7−i.p=7-i.p=7−i.

  3. Now compute z12+z22z_1^2+z_2^2z12​+z22​: z12+z22=(z1+z2)2−2z1z2=25−2(7−i)=11+2i.z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2=25-2(7-i)=11+2i.z12​+z22​=(z1​+z2​)2−2z1​z2​=25−2(7−i)=11+2i.

  4. Use z14+z24=(z12+z22)2−2(z1z2)2.z_1^4+z_2^4=(z_1^2+z_2^2)^2-2(z_1z_2)^2.z14​+z24​=(z12​+z22​)2−2(z1​z2​)2.

    First, (z12+z22)2=(11+2i)2=121+44i−4=117+44i.(z_1^2+z_2^2)^2=(11+2i)^2=121+44i-4=117+44i.(z12​+z22​)2=(11+2i)2=121+44i−4=117+44i.

    Next, (z1z2)2=(7−i)2=49−14i+i2=48−14i.(z_1z_2)^2=(7-i)^2=49-14i+i^2=48-14i.(z1​z2​)2=(7−i)2=49−14i+i2=48−14i. Therefore, 2(z1z2)2=96−28i.2(z_1z_2)^2=96-28i.2(z1​z2​)2=96−28i.

    So, z14+z24=(117+44i)−(96−28i)=21+72i.z_1^4+z_2^4=(117+44i)-(96-28i)=21+72i.z14​+z24​=(117+44i)−(96−28i)=21+72i.

  5. Its modulus is

    =\sqrt{441+5184} =\sqrt{5625}=75.$$

  6. Therefore, the correct option is D: 75.\boxed{\text{D: }75}.D: 75​.

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