Question 1184218
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If {{{ (3/2 + isqrt(3)/2)^50 = 3^25(x + iy) }}} where x and y are real numbers then find x and y?
Note: Here sqrt(3) means square root of 3.
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<pre>
The number  {{{3/2 + i*(sqrt(3)/2)}}}  has the modulus  r = {{{sqrt((3/2)^2 + 3/4)}}} = {{{sqrt(9/4 + 3/4)}}} = {{{sqrt(12/4)}}} = {{{sqrt(3)}}}

                        and the argument  a = {{{pi/6}}}.


THEREFORE,  {{{3/2 + i*(sqrt(3)/2)}}} = {{{sqrt(3)*cis(pi/6)}}}.


Hence,  the left side number  {{{(3/2 + isqrt(3)/2)^50}}} is equal to  {{{(sqrt(3))^50*cis(50pi/6)}}} = {{{3^25*cis(2pi/6)}}} = {{{3^25*cis(pi/3)}}}.


It implies that  x + iy = {{{cis(pi/3)}}} = {{{1/2 + i*(sqrt(3)/2)}}}.


So  x = {{{1/2}}},  y = {{{sqrt(3)/2}}}.      <U>ANSWER</U>
</pre>

Solved.