How do you solve #u^4 = 1#?

Question

Kennedy Adams · Accepted Answer

#u=1,-1,i, -i#

In essence, our equation is a difference of squares, which is expressed as

#(u^2+1)(u^2-1)#

We can solve by setting both of these to zero in order to obtain

#u^2=-1=>u=sqrt(-1)=+-i#

#u^2=1=>u=+-1#

Consequently, our solutions are

#u=1,-1,i, -i#

I hope this is useful.

Avery Alexander · Answer

Given #4#th degree equation will have four roots : #\pm1, \pmi# Given equation: #u^4=1# is a #4#th degree equation hence it will have #4# roots as follows #u^4=1# #u^4=e^{i(0)}# #u^4=e^{i(2k\pi)}# #u=(e^{i2k\pi})^{1/4}# #u=e^{i1/4\cdot 2k\pi}# #u=e^{i{k\pi}/2}# #u=\cos({k\pi}/2)+i\sin({k\pi}/2)# Where, #k# is an integer such that #k=0, 1, 2, 3# Now, setting the values of #k# ,as #k=0, 1, 2 , 3# in above general solution, we get all four roots of given #4#th degree equation as follows #u_1=\cos({(0)\pi}/2)+i\sin({(0)\pi}/2)=1# #u_2=\cos({(1)\pi}/2)+i\sin({(1)\pi}/2)=i# #u_3=\cos({(2)\pi}/2)+i\sin({(2)\pi}/2)=-1# #u_4=\cos({(3)\pi}/2)+i\sin({(3)\pi}/2)=-i# hence, all four roots are #\pm1, \pmi#

Genesis Allen · Answer

Answer below

#u^4 -1 = 0 # #=> (u^2+1)(u^2-1) = 0 # #=> u^2-1 = 0 => u = { -1 , 1 } # #=> u^2 + 1= 0 => u = { i , - i } # #therefore u = { -1 , 1 , -i , i } #

Jace Anderson · Answer

undefined To solve the equation $ u^4 = 1 $, you can take the fourth root of both sides. This gives you $ u = \pm 1 $. Therefore, the solutions to the equation are $ u = 1 $ and $ u = -1 $.