If z1 = sqrt(3) + isqrt(3) and z2 = sqrt(3) + i, then the complex number (z1/z2)^50 lies in the

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q8 – AP-EAMCET 2026

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Home Maths Complex & QuadraticQuestion #62

z_{1} = \sqrt{3} + i \sqrt{3}

and

z_{2} = \sqrt{3} + i

, then the complex number

{(\frac{z_{1}}{z_{2}})}^{50}

lies in the

Clue 1 of 3: Polar Form Conversion

It is generally easier to perform multiplication, division, and exponentiation of complex numbers when they are in polar form. Convert $z_{1}$ and $z_{2}$ to their polar forms first.

🧠 Full Solution Path

Step 1: Convert

z_{1}

and

z_{2}

to Polar Form✦ Active

First, convert the given complex numbers $z_{1}$ and $z_{2}$ into their polar forms, $r (\cos θ + i \sin θ)$ , where $r$ is the modulus and $θ$ is the argument.

z_{1} = \sqrt{3} + i \sqrt{3}

| z_{1} | = \sqrt{(\sqrt{3})^{2} + (\sqrt{3})^{2}} = \sqrt{3 + 3} = \sqrt{6}

\arg (z_{1}) = \tan^{- 1} (\frac{\sqrt{3}}{\sqrt{3}}) = \tan^{- 1} (1) = \frac{π}{4}

z_{1} = \sqrt{6} (\cos \frac{π}{4} + i \sin \frac{π}{4})

z_{2} = \sqrt{3} + i

| z_{2} | = \sqrt{(\sqrt{3})^{2} + (1)^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2

\arg (z_{2}) = \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6}

z_{2} = 2 (\cos \frac{π}{6} + i \sin \frac{π}{6})

💡 Teacher's Secret Hint

Always ensure the argument $θ$ is in the correct quadrant based on the signs of the real and imaginary parts of the complex number.

Step 2: Calculate the Quotient

\frac{z_{1}}{z_{2}}

in Polar Form○ Expand

Use the property for division of complex numbers in polar form: $\frac{r_{1} (\cos θ_{1} + i \sin θ_{1})}{r_{2} (\cos θ_{2} + i \sin θ_{2})} = \frac{r_{1}}{r_{2}} (\cos (θ_{1} - θ_{2}) + i \sin (θ_{1} - θ_{2}))$ . Calculate the difference of the arguments:

\frac{z_{1}}{z_{2}} = \frac{\sqrt{6}}{2} (\cos (\frac{π}{4} - \frac{π}{6}) + i \sin (\frac{π}{4} - \frac{π}{6}))

\frac{π}{4} - \frac{π}{6} = \frac{3 π - 2 π}{12} = \frac{π}{12}

\frac{z_{1}}{z_{2}} = \frac{\sqrt{6}}{2} (\cos \frac{π}{12} + i \sin \frac{π}{12})

💡 Teacher's Secret Hint

Remember that when dividing complex numbers, you divide the moduli and subtract the arguments.

Step 3: Apply De Moivre's Theorem to find

{(\frac{z_{1}}{z_{2}})}^{50}

○ Expand

Now, apply De Moivre's Theorem to raise the complex number to the power of 50. The modulus will be raised to the power of 50, and the argument will be multiplied by 50.

{(\frac{z_{1}}{z_{2}})}^{50} = {(\frac{\sqrt{6}}{2})}^{50} (\cos (50 \cdot \frac{π}{12}) + i \sin (50 \cdot \frac{π}{12}))

Simplify the argument: 50 \cdot \frac{π}{12} = \frac{25 π}{6}

{(\frac{z_{1}}{z_{2}})}^{50} = {(\frac{\sqrt{6}}{2})}^{50} (\cos (\frac{25 π}{6}) + i \sin (\frac{25 π}{6}))

💡 Teacher's Secret Hint

The modulus ${(\frac{\sqrt{6}}{2})}^{50}$ is a positive real number and only affects the distance from the origin, not the quadrant. The quadrant is determined solely by the argument.

Step 4: Determine the Quadrant○ Expand

To determine the quadrant, we need to find the principal value of the argument $\frac{25 π}{6}$ . We can do this by subtracting multiples of $2 π$ until the angle lies in the interval $[0, 2 π)$ .

\frac{25 π}{6} = \frac{24 π + π}{6} = 4 π + \frac{π}{6}

Since 4 π represents two full rotations, the effective argument is \frac{π}{6} .

The angle θ = \frac{π}{6} satisfies 0 < θ < \frac{π}{2} .

Therefore, the complex number ${(\frac{z_{1}}{z_{2}})}^{50}$ lies in the First Quadrant.

💡 Teacher's Secret Hint

Angles in the first quadrant are between $0$ and $π / 2$ radians (or $0^{\circ}$ and $90^{\circ}$ ). Angles in the second quadrant are between $π / 2$ and $π$ , third between $π$ and $3 π / 2$ , and fourth between $3 π / 2$ and $2 π$ .

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Maths Question 8 – AP-EAMCET 2026

Clue 1 of 3: Polar Form Conversion

🧠 Full Solution Path