Question 1102471: The equation 𝑧^2 = −5 + 12𝑖 has two (complex) solutions. Note that
𝑧 = sqrt (−5 + 12𝑖)is not a particularly useful way to write a solution, since it does not tell us the real and imaginary parts of the solution. We are actually trying to find solutions in the form 𝑧 = 𝑥 + 𝑖𝑦. Show that 𝑧 = 2 + 3𝑖 is such a solution and find the second solution. I know how to prove that z= 2 +3i is a solution, but I just don't know how to find the other solution? Please explain!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 𝑧^2 = −5 + 12𝑖
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To solve this problem, we use the polar form of a complex number and De Moivre's Theorem
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The polar form of a complex number(a+bi, where a,b are real numbers) is r(cos theta + i * sin theta) where r > or = to zero, r is the distance formula = square root(a^2+b^2) and tan theta = b/a
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write 𝑧^2 = −5 + 12𝑖 in polar form
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r = square root((-5)^2 + 12^2) = square root(25 + 144) = 13
tan theta = 12/-5 = -2.4 and theta is -67.38
Note that 180 - 67.38 is 112.62
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then z^2 = 13(cos (112.62) + i * sin(112.62))
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using De Moivre's Theorem, we know that z^2 has two roots
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let n be a positive number(2 for this problem) and let w(k) be the kth root, then
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w(k) = r^(1/2)[ cos((112.62 + 2kpi)/2) + ( i * sin((112.62 + 2kpi/2)))
where k = 0, 1 an pi = 180 degrees
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w(0) = 13(1/2) * (cos(112.62 / 2) + (i * sin(112.62 / 2))
w(0) = 2 + 3i
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w(1) = 13(1/2) * (cos((112.62+360) / 2) + (i * sin((112.62+360) / 2))
w(1) = -2 -3i
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the other solution is -2 -3i
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