document.write( "Question 1071957: Find all the complex solutions of the equation.
\n" ); document.write( "x^4 + i = 0 \n" ); document.write( "
Algebra.Com's Answer #686904 by math_helper(2461)\"\" \"About 
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Find all the complex solutions of the equation.
\n" ); document.write( "x^4 + i = 0
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\n" ); document.write( "\n" ); document.write( "\"+x%5E4+=+-i+\"
\n" ); document.write( "Use the fact that -i = e^(3(pi)i/2) :
\n" ); document.write( " x^4 = e^(3(pi)i/2)
\n" ); document.write( "Here's where its a little tricky - to find all the solutions, we must account for the fact that
\n" ); document.write( "e^w = e^(w+2(pi)k) for k=0,1,2,… (i.e. the exponential aliases on top of itself every 2(pi))
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\n" ); document.write( "Re-writing:
\n" ); document.write( " x^4 = e^((3(pi)/2) + 2k(pi))i
\n" ); document.write( "Raise both sides to the 1/4 power:
\n" ); document.write( " (x^4)^(1/4) = e^(((3(pi)/2) + 2k(pi))i * (1/4))
\n" ); document.write( " x = e^((3(pi)/8 + k*(pi)/2)i)
\n" ); document.write( "--
\n" ); document.write( "Using Euler's equation: e^(ni) = cos (n) + i*sin(n), we re-write the above:
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\n" ); document.write( " x = cos((3(pi)/8 + k(pi)/2) + i*sin(3(pi)/8 + k(pi)/2)\r
\n" ); document.write( "\n" ); document.write( "k=0: x = cos(3(pi)/8) + i*sin(3(pi)/8)
\n" ); document.write( " [ approx: x = 0.38268343 + i*0.92387953 ]
\n" ); document.write( "—
\n" ); document.write( "k=1: x = cos(3(pi)/8 + (pi)/2) + i*sin(3(pi)/8 + (pi)/2)
\n" ); document.write( " [ approx: x = -0.92387953 + i*0.38268343 ]
\n" ); document.write( "—
\n" ); document.write( "k=2: x = cos(3(pi)/8 + (pi)) + i*sin(3(pi)/8 + (pi))
\n" ); document.write( " [ approx: x = -0.38268343 - i*0.92387953 ]
\n" ); document.write( "—
\n" ); document.write( "k=3: x = cos(3(pi)/8 + 3(pi)/2) + i*sin(3(pi)/8 + 2(pi)/2)
\n" ); document.write( " [ approx: x = 0.92387953 - i*0.38268343 ]
\n" ); document.write( "—
\n" ); document.write( "When k reaches 4, the 2nd term in cos( ) and sin( ) reaches 2(pi) which means we've wrapped around once so the above four answers are the unique solutions (the rest are aliases). \r
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