Complex Numbers — JEE Main practice

1. Recognize that $𝑥^2+𝑥+1=0$ means $𝑥$ is a non-real cube root of unity $\omega$, since $\omega$ satisfies exactly this equation.

2. Recall the key property of cube roots of unity: $\omega^k+\frac{1}{\omega^k}$ depends on whether $𝑘$ is a multiple of 3. $k=3n \ \Rightarrow\ \omega^k+\frac{1}{\omega^k}=2, \qquad k\ne3n \ \Rightarrow\ \omega^k+\frac{1}{\omega^k}=-1$

3. Among $k=1$ to $25$, count how many are multiples of 3 and how many are not. $\text{Multiples of 3 } (k=3,6,\dots,24): 8 \text{ values}$ $\text{Remaining: } 25-8=17 \text{ values}$

4. Raise each case to the 4th power, since the sum required is of 4th powers: multiples of 3 contribute $2^4=16$ each, others contribute $(-1)^4=1$ each. $\text{Sum} = 8\times16 + 17\times1$

5. Compute the total. $= 128+17 = 145$

Hence, the answer is 145, option B.

1. The condition $|𝑧-5|\le3$ describes a filled disk centered at $(5,0)$ with radius 3. The point with maximum positive argument lies on the boundary, where the line from the origin is tangent to the circle.

2. For a circle of radius $r$ whose center is at distance $d$ from the origin, the tangent line from the origin touches it where $\sin\theta=\frac{r}{d}$, giving the maximum argument $\theta$. $\sin\theta = \frac{3}{5} \ \Rightarrow\ \cos\theta = \frac45$

3. The tangency point lies at distance $\sqrt{d^2-r^2}=4$ from the origin (right triangle formed by origin, center, and tangent point), so $|z|=4$.

4. Write $𝑧$ using its modulus and argument. $z = 4\left(\frac45+i\frac35\right) = \frac{16}{5}+\frac{12i}{5} \ \Rightarrow\ 5z=16+12i$

5. Substitute $5z$ into the target expression. $\frac{5z-12}{5zi+16} = \frac{(16+12i)-12}{(16+12i)i+16} = \frac{4+12i}{4+16i} = \frac{1+3i}{1+4i}$

6. Take the modulus of this ratio using $|a+bi|=\sqrt{a^2+b^2}$ for numerator and denominator. $\left|\frac{1+3i}{1+4i}\right| = \frac{\sqrt{10}}{\sqrt{17}}$

7. Square this and multiply by 34 as required. $34\cdot\frac{10}{17} = 20$

Hence, the answer is 20, option A.

1. Write $𝑧=𝑥+i𝑦$ and expand $4𝑧^2+\bar 𝑧=0$ in terms of $𝑥,𝑦$. $4(x+iy)^2+x-iy=0 \ \Rightarrow\ (4x^2-4y^2+x)+i(8xy-y)=0$

2. Separate into real and imaginary parts, since both must vanish. $4x^2-4y^2+x=0, \qquad y(8x-1)=0$

3. From the second equation, either $y=0$ or $x=\frac18$. Consider $y=0$ first. $4x^2+x=0 \ \Rightarrow\ x=0 \text{ or } x=-\frac14$

4. This gives two solutions: $z_1=0$ and $z_2=-\frac14$. Compute their squared moduli. $|z_1|^2=0, \qquad |z_2|^2=\frac{1}{16}$

5. Now consider $x=\frac18$, substituting into the first equation to solve for $𝑦$. $4\cdot\frac{1}{64}-4y^2+\frac18=0 \ \Rightarrow\ 4y^2=\frac{3}{16} \ \Rightarrow\ y=\pm\frac{\sqrt3}{8}$

6. This gives two more solutions with equal squared moduli. $z_{3,4} = \frac18\pm\frac{\sqrt3}{8}i \ \Rightarrow\ |z_3|^2=|z_4|^2=\frac{1}{64}+\frac{3}{64}=\frac{1}{16}$

7. Sum all four squared moduli. $0+\frac{1}{16}+\frac{1}{16}+\frac{1}{16} = \frac{3}{16}$

Hence, the answer is $\frac{3}{16}$, option D.