If \[i{z^4} + 1 = 0\], then $z$ can take the value of
A. $\frac{{1 + i}}{{\sqrt 2 }}$
B. $\cos \frac{\pi }{8} + i\sin \frac{\pi }{8}$
C. $\frac{1}{{4i}}$
D. $i$
Answer
363.9k+ views
Hint: To find the value of $z$ we are going to simplify the given equation and bring it in terms of $z$.
Complete step by step answer:
The given equation can be written as,
$i{z^4} = - 1$
${z^4} = - \frac{1}{i}$
${z^4} = i$
If we simplify the above equation, we get,
$z = {\left( i \right)^{1/4}} = {\left( {0 + i} \right)^{^{1/4}}}$
Option B is the correct answer for this question.
Note: In the above equation we added zero in the front to get it into the general form of complex number.
Therefore,
$z = {\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)^{1/4}}$
$z = \cos \frac{\pi }{8} + i\sin \frac{\pi }{8}$
Complete step by step answer:
The given equation can be written as,
$i{z^4} = - 1$
${z^4} = - \frac{1}{i}$
${z^4} = i$
If we simplify the above equation, we get,
$z = {\left( i \right)^{1/4}} = {\left( {0 + i} \right)^{^{1/4}}}$
Option B is the correct answer for this question.
Note: In the above equation we added zero in the front to get it into the general form of complex number.
Therefore,
$z = {\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right)^{1/4}}$
$z = \cos \frac{\pi }{8} + i\sin \frac{\pi }{8}$
Last updated date: 23rd Sep 2023
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Total views: 363.9k
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