AP EAPCET 21 May 2025 Shift 1 Paper

z1 and z2 are both nth roots of unity. This means each one can be written in the form e‌

i2πr

n

, where r is an integer.
Let z1=e‌

i2πr1

n

and z2=e‌

i2πr2

n

for some integers r1 and r2.
If the line segment joining z1 and z2 makes a right angle at the origin, this means the points z1,z2, and the origin form a right-angled triangle with the right angle at the origin.
This can only happen if ‌

z1

z2

is a purely imaginary number. A complex number is purely imaginary when its argument (angle) is ‌

π

2

or ‌

3π

2

, or in general, ‌

π

2

+mπ where m is an integer.
We find: ‌

z1

z2

=e‌

i2π

n

(r1−r2)
For this to be purely imaginary, we need: ‌

2π

n

(r1−r2)=‌

π

2

+mπ
Divide both sides by π:‌

2

n

(r1−r2)=‌

2m+1

2

Now solve for n:n=‌

4(r1−r2)

2m+1

Since n must be a positive integer, 2m+1 must divide evenly into 4(r1−r2). The simplest way for this to always work for integer values is when n is a multiple of 4 . Therefore, n=4k for some positive integer k.