Question 1181165
.
Use De Moivre's Theorem to show that integral powers of (-1 + i)/(√2) are real, and which are imaginary
~~~~~~~~~~~~~~~~~~~~~


<pre>
The complex number  z = {{{(-1+i)/sqrt(2)}}}  has the modulus 1  and the argument  {{{3pi/4}}}.


So, in cis-form,  z = {{{1*cis(3pi/4)}}} = {{{cis(3pi/4)}}}.


It means that z itself and all integer degrees of z have the modulus 1, i.e. lie on a unit circle 
in complex plane.


According to De Moivre's theorem, the degrees of z are


    {{{z^1}}} = {{{cis(3pi/4)}}}

    {{{z^2}}} = {{{cis(6pi/4)}}} = {{{cis(3pi/2)}}} = -i   (pure imaginary)

    {{{z^3}}} = {{{cis(9pi/4)}}}

    {{{z^4}}} = {{{cis(12pi/4)}}} = {{{cis(3pi)}}} = -1    (real number)
    
    {{{z^5}}} = {{{cis(15pi/4)}}}

    {{{z^6}}} = {{{cis(18pi/4)}}} = {{{cis(9pi/2)}}} = i   (pure imaginary)

    {{{z^7}}} = {{{cis(21pi/4)}}}

    {{{z^8}}} = {{{cis(24pi/4)}}} = {{{cis(6pi)}}} = 1     (real number)


The degrees of z that follow after  {{{z^8}}},  repeat these numbers cyclically


    {{{z^9}}} = {{{z^1}}}  

    {{{z^10}}} = {{{z^2}}} = -i  (imaginary)

    {{{z^11}}} = {{{z^3}}}

    {{{z^12}}} = {{{z^4}}} = -1  (real number)



    {{{z^13}}} = {{{z^5}}}  

    {{{z^14}}} = {{{z^6}}} =  i  (imaginary)

    {{{z^15}}} = {{{z^7}}}

    {{{z^16}}} = {{{z^8}}} =  1  (real number)


So, the pattern is this:  {{{z^n}}}  is real     if and only n is of the form  n = 4k  (i.e. n is a multiple of 4), and

                          {{{z^n}}}  is pure imaginary if and only n is of the form  n = 4k+2  (i.e. n gives the remainder of 2 when is divided by 4).



<U>ANSWER</U>.  {{{z^n}}}  is real if and only if  n == 0  mod 4;

         {{{z^n}}}  is pure imaginary if and only if  n == 2  mod 4.
</pre>

Solved.