SOLUTION: Find all complex numbers z such that z^2 = 2i. Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (

Algebra ->  Real-numbers -> SOLUTION: Find all complex numbers z such that z^2 = 2i. Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (      Log On


   

Question 1028076: Find all complex numbers z such that z^2 = 2i.
Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (Don't include quotes in your answer.)
Note: This problem is not about functions.
Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find all complex numbers z such that z^2 = 2i.
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z^2 = 2cis90
z = sqrt(2)cis(45), sqrt(2)cis(225)
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z = sqrt(2)*sqrt(2)/2 + isqrt(2)*sqrt(2)/2
= 1 + i
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z = sqrt(2)cis(225)
sqrt(2)*(-sqrt(2)/2) + i*sqrt(2)*(-sqrt(2)/2)
= -1 - i

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Use the fact that

Since you can conclude that the magnitude of z is .

Then you can conclude that and are the solutions. Since the equation is quadratic, it must have at most two complex solutions, including multiple roots.