Question

How do you find all solutions to ${{x}^{4}}+(1+i)=0$ ?
(a) By solving the equation
(b) Using de moivre’s identity
(c) Using complex analysis
(d) All of the above

Answer 1

Hint: This is not a simple complex equation which can be solved with the help of middle term and factorization to start with. So, to solve this equation we are going to use the de moivre’s identity and get the solution we needed.

Complete step-by-step answer:
According to the question, we are given the equation, ${{x}^{4}}+(1+i)=0$
$\Rightarrow {{x}^{4}}=-(1+i)$
Using de moivre’s identity,
${{e}^{i\phi }}=\cos \phi +i\sin \phi $ , we have,
\[-(1+i)=-\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right)=-\sqrt{2}\left( \cos \phi +i\sin \phi \right)\]
Thus, we have the given form of the given equation.
For de Moivre's theorem, we first gain some intuition for de Moivre's theorem by considering what happens when we multiply a complex number by itself. This shows that by squaring a complex number, the absolute value is squared and the argument is multiplied by 2. For, n greater than and equal to 3, de Moivre's theorem generalizes this to show that to raise a complex number to the nth power, the absolute value is raised to the nth power and the argument is multiplied by n.
So, we get, by comparing both sides, $\phi =\dfrac{\pi }{4}+2k\pi ,k=0,\pm 1,\pm 2,...$
But, here we again can see,\[-\sqrt{2}\left( \cos \phi +i\sin \phi \right)=\sqrt{2}{{e}^{i\pi }}{{e}^{i\left( \dfrac{\pi }{4}+2k\pi \right)}}=\sqrt{2}{{e}^{i\left( \dfrac{5\pi }{4}+2k\pi \right)}}\]
And as, we have, ${{e}^{i\pi }}+1=0$ , because we know that the value of $\cos \pi =-1$ and $\sin \pi =0$,
We now get, ${x^4} = \sqrt 2 {e^{i\left( {\dfrac{{5\pi }}{4} + 2k\pi } \right)}}$
$\Rightarrow x=\sqrt[8]{2}{{e}^{i\left( \dfrac{5\pi }{16}+k\dfrac{\pi }{2} \right)}},\forall k=0,\pm 1,\pm 2,....$

So, the correct answer is “Option (b)”.

Note: The De Moivre’s formula is important because it connects complex numbers and trigonometry by expanding the left hand side. Then comparing the real and imaginary parts under the assumption that x is real, it is possible to derive useful expressions for cos nx and sin nx in terms of cos x and sin x. It is quite an important formula and identity for solving any algebra problem.