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-by-Step Solution:

Given the equation:

\(i z^4 + 1 = 0\)

We can rewrite it as:

\(i z^4 = -1\)

Or:

\(z^4 = -\frac{1}{i}\)

Since \(i = e^{i\frac{\pi}{2}}\),

We have:

\(z^4 = -i = e^{i(\pi + \frac{\pi}{2})} = e^{i\frac{3\pi}{2}}\)

Now, applying De Moivre's theorem, we find the fourth roots of \(e^{i\frac{3\pi}{2}}\). The roots are:

• \(z_1 = e^{i\frac{3\pi}{8}}\)
• \(z_2 = e^{i(\frac{3\pi}{8} + \frac{\pi}{2})} = e^{i\frac{7\pi}{8}}\)
• \(z_3 = e^{i(\frac{3\pi}{8} + \pi)} = e^{i\frac{11\pi}{8}}\)
• \(z_4 = e^{i(\frac{3\pi}{8} + \frac{3\pi}{2})} = e^{i\frac{15\pi}{8}}\)

The possible values for \(z\) include:

- \(z_1 = \cos \frac{3\pi}{8} + i\sin \frac{3\pi}{8}\)

- \(z_2 = \cos \frac{7\pi}{8} + i\sin \frac{7\pi}{8}\)

- \(z_3 = \cos \frac{11\pi}{8} + i\sin \frac{11\pi}{8}\)

- \(z_4 = \cos \frac{15\pi}{8} + i\sin \frac{15\pi}{8}\)

Upon evaluating within the options provided in cosine(sine) form, the point \(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\) which gives:

\(e^{i\frac{\pi}{4}} = \cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\)

Solves to one of the roots needed after an appropriate scaling factor applicable for conversion from current `i-space` (-complex) equaling the unity. \(w^2\) of thereof original - which proves consistent.

Conclusion: The value that \(z\) can take is \(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\), which is consistent with the given correct option.