Given $(x+iy)^2 = 8+6i$, find the values of $x$ and $y$. Hence find $\sqrt{8+6i}$.
My question is when we solve we get $x = 3$ and $x = -3$, which give and $y = 1$ and $y = -1$ then Why $-3-i$ is not the solution of $\sqrt{8+6i}$ ?
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2What is $(-3-i)^2$ – lab bhattacharjee Dec 12 '13 at 16:02
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Possible duplicate of How do I get the square root of a complex number? – gebruiker May 03 '16 at 08:59
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To check the solution, we have to square $z$ and ideally get the RHS($8+6i)$ $(-3-i)(-3-i)=9+3i+3i+i^2=6i+9-1=8+6i$ as needed So, $-3-i$ is in fact a solution and is equal to $\sqrt{8+6i}$
John
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Thanks for the clarification John, I had the confusion because both the answer in the back of the book and wolfram alpha did't show (-3-6i) – user115368 Dec 29 '13 at 14:50
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@user115368 It should be noted that the square root function is not well defined for complex numbers. Wolfram Alpha and perhaps your book were giving you the principal root-which has the positive real component. You may find more information asked by another person here: http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number – John Dec 29 '13 at 16:32