Question

If \[i{{z}^{4}}+1=0,\] 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: In this question, we will first write the value of z in terms of i and then solve it further by using De – Moivre’s theorem and by substituting various values to the power of i.

Complete step-by-step answer:
We are given the equation: \[i{{z}^{4}}+1=0\] where we have to find the value of z. For writing this equation as the value of z, we will perform the following steps of simplification.
\[i{{z}^{4}}+1=0\]
Adding – 1 on both the sides of the equation we get,
\[\Rightarrow i{{z}^{4}}=-1\]
Dividing by i into both sides we get,
\[\Rightarrow {{z}^{4}}=\dfrac{-1}{i}\]
We know that,
\[{{i}^{2}}=-1\]
Therefore, substituting the value of – 1 as \[{{i}^{2}},\] we get,
\[\Rightarrow {{z}^{4}}=\dfrac{{{i}^{2}}}{i}\]
\[\Rightarrow {{z}^{4}}=i\]
Now, this value can be written in the form of a complex number by putting the real number as 0 and the imaginary number as 1.
\[\therefore {{z}^{4}}=0+1i\]
Now, let’s convert the above equation into a polar form.
We know that,
\[z=\left( \cos \theta +i\sin \theta \right)\]
Also, \[\cos \dfrac{\pi }{2}=0;\sin \dfrac{\pi }{2}=1\]
Therefore substituting 0 as \[\cos \dfrac{\pi }{2}\] and 1 as \[\sin \dfrac{\pi }{2},\] we get,
\[{{z}^{4}}=\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)\]
\[\Rightarrow z={{\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)}^{\dfrac{1}{4}}}\]
By using De –Moivre’s theorem which states that, \[{{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta \]
We get, \[\theta =\dfrac{\pi }{2};n=\dfrac{1}{4}\]
On substitution, we get,
\[z={{\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)}^{\dfrac{1}{4}}}\]
\[z=\left( \cos \left( \dfrac{\pi }{2}.\dfrac{1}{4} \right)+i\sin \left( \dfrac{\pi }{2}.\dfrac{1}{4} \right) \right)\]
\[z=\cos \dfrac{\pi }{8}+i\sin \dfrac{\pi }{8}\]
Hence, option (b) is the right answer.

Note: You can guess that we have to use De – Moivre’s theorem by looking at the option. Another way to crack this question is to work through the options and try to find the simplified form of the question through it to get the answer. Also, remember the values of the powers of i:
\[i=\sqrt{-1};\text{ }{{i}^{2}}=-1;\text{ }{{i}^{3}}=-i;\text{ }{{i}^{4}}=i\]