SOLUTION: how do you convert (1+i)/((square root of 3)+i) to trigonometric form?

Question 383482: how do you convert (1+i)/((square root of 3)+i) to trigonometric form?

Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
how do you convert (1+i)/((square root of 3)+i) to trigonometric form?

(1 + i)/(sqrt(3) + i)
need to multiply the numerator and denominator
by the conjugate of the denominator
sqrt(3) + i conjugate is sqrt(3) - i
multiply complex numbers with FOIL (First Outer Inner Last)
i^2 = -1
numerator: (1 + i)(sqrt(3) - i) = sqrt(3) - i + sqrt(3)i - i^2
sqrt(3) + 1 + (sqrt(3) - 1)i
denominator: (sqrt(3) + i)(sqrt(3) - i) = 3 - i^2 = 3 + 1 = 4
divided:
(1 + i)/(sqrt(3) + i) = (sqrt(3) + 1)/4 + ((sqrt(3) - 1)/4)i
complex numbers in standard form are a + bi,
so answer approximately with a and b rounded to 6 digits is:
0.683013 + 0.183013i
trigonometric (polar) form:
complex numbers when plotted the imaginary part is the y-axis
and the real part is the x-axis, quadrants numbered counter-clockwise,
here the real and imaginary part are both positive so complex number is
in quadrant 1
and of form z = r(cos(theta) + isin(theta))
r = sqrt(x^2 + y^2) = sqrt(0.683013^2 + 0.183013^2)
r = sqrt(0.5) = 1/(sqrt(2)) = sqrt(2)/2 = 0.707107 rounded to 6 places
tan (theta) = y/x = ((sqrt(3) - 1)/4) / ((sqrt(3) + 1)/4)
tan (theta) = (sqrt(3) - 1)/(sqrt(3) + 1)
tan (theta) = ((sqrt(3) - 1)(sqrt(3) - 1))/((sqrt(3) + 1)(sqrt(3) - 1))
tan (theta) = (3 - 2sqrt(3) + 1)/(3 - 1)
(multiplied top and bottom by (sqrt(3) - 1))
tan (theta) = (4 - 2sqrt(3))/2 = 2 - sqrt(3) = 0.267949 to 6 places
theta = 15 degrees
so:
sqrt(3) + 1)/4 + ((sqrt(3) - 1)/4)i or 0.683013 + 0.183013i is
(sqrt(2)/2)cos(15) + (sqrt(2)/2)sin(15)i

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