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Complex Numbers

Section: Mathematics

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Question : 16 of 50

Marks: +1, -0

If

\left(\frac{\cos\theta + i\sin\theta}{\sin\theta + i\cos\theta}\right)^{2024} + \left(\frac{1+\cos\theta + i\sin\theta}{1-\cos\theta + i\sin\theta}\right)^{2025} = x+iy

, then the value of

x+y

\theta = \frac{\pi}{2}

[AP EAPCET 21 May 2025 S2]

1
-1
2
2024

Solution:

We have,

\left(\frac{\cos\theta + i\sin\theta}{\sin\theta + i\cos\theta}\right)^{2024} + \left(\frac{1+\cos\theta + i\sin\theta}{1-\cos\theta + i\sin\theta}\right)^{2025}

= \left\{ \frac{\cos\theta + i\sin\theta}{i(\cos\theta - i\sin\theta)} \right\}^{2024} + \left( \frac{2\cos^2\frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \right)^{2025}

= \left( \frac{e^{i\theta}}{e^{-i\theta}} \right)^{2024} + \frac{ \left(2\cos\frac{\theta}{2}\right)^{2025} \left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)^{2025} }{ \left(2\sin\frac{\theta}{2}\right)^{2025} \left(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2}\right)^{2025} }

= (e^{2i\theta})^{2024} + \left(\cot\frac{\theta}{2}\right)^{2025} \frac{ (e^{i\theta/2})^{2025} }{ i^{2025} (e^{-i\theta/2})^{2025} }

= (e^{2i\theta})^{2024} + \left(\cot\frac{\theta}{2}\right)^{2025} \frac{1}{i} (e^{i\theta})^{2025}

= (\cos 2\theta + i\sin 2\theta)^{2024} - \left(\cot\frac{\theta}{2}\right)^{2025} i(\cos\theta + i\sin\theta)^{2025}

Now, put

\theta = \frac{\pi}{2}

= (-1)^{2024} - i(i)^{2025} = 1 - i^2 = 1 + 1 = 2

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