document.write( "Question 115817: change -sqrt3 + i in trigonometric form and find\r
\n" ); document.write( "\n" ); document.write( "1. (-sqrt3+i)^5, answer in rectangular form such as a+bi?\r
\n" ); document.write( "\n" ); document.write( "2. (-sqrt3 + i)^1/3, write all answers separately in trigonometric form using degree angles?
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Algebra.Com's Answer #84269 by stanbon(75887)\"\" \"About 
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change -sqrt3 + i in trigonometric form and find
\n" ); document.write( "1. (-sqrt3+i)^5, answer in rectangular form such as a+bi?
\n" ); document.write( "r = sqrt[(-sqrt3)^2 + 1^2] = sqrt[4] = 2
\n" ); document.write( "theta = tan^-1[1/(-srt3)] = 150 degrees
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\n" ); document.write( "Therefore: (-sqrt3+i) = [2cis(150)]^5
\n" ); document.write( "= 32cis(5*150)
\n" ); document.write( "= 32 cis (750)
\n" ); document.write( "= 32cis(30)
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\n" ); document.write( "2. (-sqrt3 + i)^1/3, write all answers separately in trigonometric form using degree angles?
\n" ); document.write( "(-sqrt3 + i)^(1/3) = [2cis(150)]^(1/3)
\n" ); document.write( "= 2^(1/3)cis((1/3)150)
\n" ); document.write( "=2^(1/3)cis(50)
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\n" ); document.write( "I'm assuming you want the principal root, not all three cube roots
\n" ); document.write( "If you want all three roots you need to use (1/3)[150+360k) where
\n" ); document.write( "k = 0, then k=1, then k=2
\n" ); document.write( "You would have angles of 50, 50+120, and 50+240
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan
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