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Maths Question 4 – JEE-MAIN 2025

Let zC be such that z2+3iz2+i=2+3i. Then the sum of all possible values of z2 is

Begin by clearing the denominator to transform the given equation into a standard polynomial form.

Step 1: Rearrange the equation into a quadratic form✦ Active

The given equation is z2+3iz2+i=2+3i. Multiply both sides by (z2+i) to clear the denominator:

z2+3i=(2+3i)(z2+i) z2+3i=2z4+2i+3iz6i+3i2 z2+3i=2z4+2i+3iz6i3 z2+3i=(2+3i)z74i

Rearrange the terms to form a quadratic equation in z:

z2(2+3i)z+(7+7i)=0

This is a quadratic equation az2+bz+c=0 with a=1, b=(2+3i), and c=(7+7i). Note that z2i for the original expression to be defined. We can verify that z=2i is not a root of the quadratic equation, so both roots will be valid.

Step 2: Apply Vieta's formulas for roots z1,z2○ Expand

Let z1 and z2 be the roots of the quadratic equation z2(2+3i)z+(7+7i)=0. According to Vieta's formulas:

z1+z2=ba=(2+3i)1=2+3i z1z2=ca=7+7i1=7+7i
Step 3: Calculate the sum of squares of the roots○ Expand

The sum of all possible values of z2 is z12+z22. We use the algebraic identity z12+z22=(z1+z2)22z1z2:

z12+z22=(2+3i)22(7+7i) z12+z22=(22+2(2)(3i)+(3i)2)(14+14i) z12+z22=(4+12i+9i2)(14+14i) z12+z22=(4+12i9)(14+14i) z12+z22=(5+12i)(14+14i) z12+z22=514+12i14i z12+z22=192i
💡 Teacher's Secret Hint

Ensure careful expansion of complex number squares and distribution of multiplication.

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