UPSC NDA I 2025 Mathematics Paper

The key concept involves simplifying complex numbers in the polar form. We use the argument (angle) of the complex number and De Moivre's theorem, which states:
(r(cos‌θ+isin‌θ))n=rn(cos(nθ)+isin‌(nθ))
Calculation:
⇒(‌

√3+i

√3−i

)3
Multiply the numerator and the denominator by the conjugate of the denominator
‌⇒‌

√3+i

√3−i

×‌

√3+i

√3+i

‌⇒‌

(√3+i)2

(√3−i)(√3+i)

‌‌

2+2√3i

4

=‌

1+√3i

2

Convert to polar form. The modulus r is
r=√‌(‌

1

2

)2+(‌

√3

2

)2=√‌‌

1

4

+‌

3

4

=√‌1=1
The argument θ is
θ=tan−1(‌

‌ √3 2

1 2

)=tan−1(√‌3)=‌

π

3

So the polar form of the complex number is
1(cos‌

π

3

+isin‌‌

π

3

)
To cube the complex number, we use De Moivre's Theorem
In our case, r=0 so,
(cos‌

π

3

+isin‌‌

π

3

)3=cos‌π+isin‌π=−1+0i=−1