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

3 Answers
Jul 21, 2018

Answer:

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

Explanation:

Our equation is essentially a difference of squares which can be rewritten as

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

To solve, we can set both of these equal to zero to get

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

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

Therefore, our solutions are

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

Hope this helps!

Answer:

Given #4#th degree equation will have four roots : #\pm1, \pmi#

Explanation:

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#

Jul 21, 2018

Answer:

Answer below

Explanation:

#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 } #