Question

If an equation is given by $i{{z}^{4}}+1=0$, then z can take the value
(a) $\dfrac{1+i}{\sqrt{2}}$
(b) $\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}$
(c) $\dfrac{1}{4i}$
(d) i

Answer 1

Hint: Find the value of ${{z}^{4}}$ from the given expression as i by using ${{i}^{2}}=-1$. Now, write i in terms of $\sin \theta \text{ and }\cos \theta \text{ as }\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}$. Now, use the Demoivre’s theorem which says that ${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta$ to find the value of z.
Complete step-by-step answer:
We are given that $i{{z}^{4}}+1=0$, we have to find the value of z. Let us consider the expression given in the question.
$i{{z}^{4}}+1=0$
By subtracting 1 from both the sides of the above equation, we get,
$i{{z}^{4}}=-1$
${{z}^{4}}=\dfrac{-1}{i}$
By multiplying i in the numerator and denominator of the RHS of the above equation, we get,
${{z}^{4}}=\dfrac{-i}{{{i}^{2}}}$
We know that $i=\sqrt{-1}\text{ or }{{i}^{2}}=-1$. By using this, we get,
${{z}^{4}}=i$
We can write the above equation as,
${{z}^{4}}=0+i\left( 1 \right)$
We know that,
$\cos \left( \dfrac{\pi }{2} \right)=0\text{ and }\sin \left( \dfrac{\pi }{2} \right)=1$
By using these, we can write the above equation as,
${{z}^{4}}=\cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right)$
By raising both the sides by the power of $\dfrac{1}{4}$, we get,
$z={{\left( \cos \dfrac{\pi }{2}+i\sin \left( \dfrac{\pi }{2} \right) \right)}^{\dfrac{1}{4}}}$
According to De Moivre’s theorem, we know that ${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta$. By using this in the above equation, we get,
$z=\cos \dfrac{1}{4}\left( \dfrac{\pi }{2} \right)+i\sin \dfrac{1}{4}\left( \dfrac{\pi }{2} \right)$
So, we get,
$z=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}$
Therefore, option (b) is the right answer.

Note: In this question, students can cross-check their answer by substituting $z=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}$ and checking if LHS = RHS as follows:
Let us consider the given equation
$i{{z}^{4}}+1=0$
By substituting $z=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}$, we get,
$i{{\left( \cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8} \right)}^{4}}+1=0$
By using Demoivre’s theorem, we get,
$i\left( \cos \dfrac{4\pi }{8}+i\sin \dfrac{4\pi }{8} \right)+1=0$
$i\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{8} \right)+1=0$
$i\left( 0+i\left( 1 \right) \right)+1=0$
${{i}^{2}}+1=0$
We know that,
${{i}^{2}}=-1$
So, we get,
– 1 + 1 = 0
0 = 0
As we have got LHS = RHS. So, our answer is correct.