Question

If \(iz^4 + 1 = 0\) then \(z\) can take the value

• A. \(\cos \frac{\pi}{8} + i\sin \frac{\pi}{8}\)
• B. \(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\)
• C. \(\cos \frac{3\pi}{8} + i\sin \frac{3\pi}{8}\)
• D. \(\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}\)

Hint:

For \(z^n = r(\cos\theta + i\sin\theta)\), roots are \(\cos\frac{\theta+2k\pi}{n}\).

Answer 1

The Correct Option is B

Step 1: Simplify the equation}
\[ iz^4 + 1 = 0 \Rightarrow iz^4 = -1 \] \[ z^4 = \frac{-1}{i} = i \] Step 2: Write in polar form}
\[ i = \cos \frac{\pi}{2} + i\sin \frac{\pi}{2} \] Step 3: Find fourth roots using De Moivre’s theorem}
\[ z = \left(\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}\right)^{1/4} \] \[ z = \cos \frac{\frac{\pi}{2} + 2k\pi}{4} + i\sin \frac{\frac{\pi}{2} + 2k\pi}{4}, \quad k=0,1,2,3 \] Step 4: Substitute values of \(k\)}
For \(k=0\): \[ z = \cos \frac{\pi}{8} + i\sin \frac{\pi}{8} \] For \(k=1\): \[ z = \cos \frac{5\pi}{8} + i\sin \frac{5\pi}{8} \] For \(k=2\): \[ z = \cos \frac{9\pi}{8} + i\sin \frac{9\pi}{8} \] For \(k=3\): \[ z = \cos \frac{13\pi}{8} + i\sin \frac{13\pi}{8} \] Step 5: Final Answer
Among the given options, the acceptable value is: \[ \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} \]