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.