Given the equation $$z^{4} = 1$$ Because equation degree is equal to = 4 - contains the even number 4 in the numerator, then the equation has two real roots. Get the root 4-th degree of the equation sides: We get: $$\sqrt[4]{z^{4}} = \sqrt[4]{1}$$ $$\sqrt[4]{z^{4}} = \left(-1\right) \sqrt[4]{1}$$ or $$z = 1$$ $$z = -1$$ We get the answer: z = 1 We get the answer: z = -1 or $$z_{1} = -1$$ $$z_{2} = 1$$
All other 2 root(s) is the complex numbers. do replacement: $$w = z$$ then the equation will be the: $$w^{4} = 1$$ Any complex number can presented so: $$w = r e^{i p}$$ substitute to the equation $$r^{4} e^{4 i p} = 1$$ where $$r = 1$$ - the magnitude of the complex number Substitute r: $$e^{4 i p} = 1$$ Using Euler’s formula, we find roots for p $$i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = 1$$ so $$\cos{\left(4 p \right)} = 1$$ and $$\sin{\left(4 p \right)} = 0$$ then $$p = \frac{\pi N}{2}$$ where N=0,1,2,3,... Looping through the values of N and substituting p into the formula for w Consequently, the solution will be for w: $$w_{1} = -1$$ $$w_{2} = 1$$ $$w_{3} = - i$$ $$w_{4} = i$$ do backward replacement $$w = z$$ $$z = w$$
The final answer: $$z_{1} = -1$$ $$z_{2} = 1$$ $$z_{3} = - i$$ $$z_{4} = i$$