Complex numbers, Find (15+8i)^(1/2) in the form of a+ib
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Complex numbers, Find (15+8i)^(1/2) in the form of a+ib

[From: ] [author: ] [Date: 13-01-21] [Hit: ]
solve by observation, substitution or otherwise and you get a=4 and b=1,√(15+8i) = 4+i-15 + 8i = √(15^2 + 8^2) * e^(iarctan(8/15)) = √17*e^(0.√(15 + 8i) = √17 * e^(0.245i) = √17cos(0.245) + i√17*sin(0.......
Find (15+8i)^(1/2) in the form of a+ib

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√(15+8i) = a+ib, so 15+8i = (a+ib)² = (a²-b²) + i(2ab)

a²-b² = 15 and ab=4, solve by observation, substitution or otherwise and you get a=4 and b=1, hence

√(15+8i) = 4+i

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15 + 8i = √(15^2 + 8^2) * e^(iarctan(8/15)) = √17*e^(0.49i)

√(15 + 8i) = √17 * e^(0.245i) = √17cos(0.245) + i√17*sin(0.245)

= 4 + i
1
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