\( \sqrt[4]{-64} = ? \) | Address Academy Question

Another Explanation (5):

উত্তর: \( \pm 2(1 \pm i) \)

প্রথমে, ধরা যাক, \( z = \sqrt[4]{-64} \). অর্থাৎ, \( z^4 = -64 \).

প্রথমে, কমপ্লেক্স সংখ্যার রূপে লিখি:

\( -64 = 64 \times (-1) \)

এবং, জানি যে, \( -1 = e^{i\pi} \) বা \( e^{i(\pi + 2k\pi)} \), যেখানে \( k \) হলো পূর্ণসংখ্যা।

অতএব, \( -64 = 64 e^{i(\pi + 2k\pi)} \), যেখানে \( k = 0,1,2,\dots \)

এবং, \( 64 = 2^6 \), সুতরাং:

\( -64 = 2^6 e^{i(\pi + 2k\pi)} \)

আমাদের লক্ষ্য হলো \( z \) such that:

\( z^4 = 2^6 e^{i(\pi + 2k\pi)} \)

তাহলে,

\( z = \sqrt[4]{2^6} \times e^{i \frac{\pi + 2k\pi}{4}} \)

এবং, \( \sqrt[4]{2^6} = 2^{6/4} = 2^{3/2} = 2 \sqrt{2} \).

সুতরাং,

\( z_k = 2 \sqrt{2} \times e^{i \frac{\pi + 2k\pi}{4}} \), যেখানে \( k=0,1,2,3 \).

প্রতিটি মূলের জন্য:

\( k=0 \): \( z_0 = 2 \sqrt{2} \times e^{i \frac{\pi}{4}} = 2 \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 2 \sqrt{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 2 \sqrt{2} \times \frac{\sqrt{2}}{2} (1 + i) = 2 \times 1 (1 + i) = 2(1 + i) \)
\( k=1 \): \( z_1 = 2 \sqrt{2} \times e^{i \frac{\pi + 2\pi}{4}} = 2 \sqrt{2} \times e^{i \frac{3\pi}{4}} = 2 \sqrt{2} (\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) = 2 \sqrt{2} \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 2 \times (-1 + i) = -2(1 - i) \)
\( k=2 \): \( z_2 = 2 \sqrt{2} \times e^{i \frac{\pi + 4\pi}{4}} = 2 \sqrt{2} \times e^{i \frac{5\pi}{4}} = 2 \sqrt{2} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) = 2 \sqrt{2} \left( -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = 2 \times (-1 - i) = -2(1 + i) \)
\( k=3 \): \( z_3 = 2 \sqrt{2} \times e^{i \frac{\pi + 6\pi}{4}} = 2 \sqrt{2} \times e^{i \frac{7\pi}{4}} = 2 \sqrt{2} (\cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}) = 2 \sqrt{2} \left( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = 2 \times (1 - i) = 2(1 - i) \)

অতএব, চারর্থমূলের সমাধানসমূহ হলো:

\( \boxed{ z = \pm 2(1 + i), \quad \pm 2(1 - i) } \)