TG EAMCET 2 May 2025 Shift 1 Paper

Letz=√3−i
|z|=√(√3)2+(−1)2=√3+1=2
And‌‌θ=tan−1(‌

−1

√3

)=−‌

π

6

or‌

11π

6

So,‌‌z=2(cos(−‌

π

6

)+isin‌(−‌

π

6

))
Now,z‌

3

7

=(√3−i)‌

3

7

First,z3=(√3−i)3
‌=23(cos(‌

−3π

6

)+isin‌(‌

−3π

6

))
‌=8(cos‌

π

2

−isin‌‌

π

2

)....(i)
Now, z‌

3

7

=(√3−i)‌

3

7

=(8)‌

1

7

(cos(‌

− π 2 +2kπ

7

)+isin‌(‌

− π 2 +2kπ

7

))....(ii)
fork=0,1,...,6
So,z3=8(0−i)=−8i‌‌[Using Eq. (i)]
Now, we know that the product ofnthroot of a complex numberAis given by(−1)n−1A, if A is a real number.
Since,n=7is odd, the product of the 7 roots of A is A itself.
∴The product of the 7th roots of A is A itself.
∴‌‌(√3−i)‌

3

7

=z3
So, product=−8i