The answer can be plus negative
z = 3 - 4i
x^2 - y^2 = 3
x^2 + y^2 = 5
2x^2 = 8, x = (+/-) 2
2y^2 = -2, y = (+/-) 1
Since Im(z) < 0, opp in sign
Therefore Answer is +/-(2-i)
If A=3-41, find sqrtA
if you let x=sqrtA, there is only one solution, which is 2-i (first degree polynomial)
However, if you use the method of squaring, so x^2 = A, there are 2 solutions. +/-(2-i) (second degree polynomial)
This also arises when you do, for example, x=sqrt4=2 but x^2=4 gives x=+/-2
for sqrt4 the answer is 2 and not -2 because you take the positive case. However, a complex number cannot be positive or negative, so what would be the correct answer to sqrtA. [the answer says +/-(2-i) :/]
The answer can be plus negative
z = 3 - 4i
x^2 - y^2 = 3
x^2 + y^2 = 5
2x^2 = 8, x = (+/-) 2
2y^2 = -2, y = (+/-) 1
Since Im(z) < 0, opp in sign
Therefore Answer is +/-(2-i)
For HSC purposes you should probably give both square roots as the answer in general (as a +/- pair).
I like the method in Terry Lee's:
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It's essentially done by inspection and showing that the inspected answer works, so I don't see why not. (It's using the well-known fact that complex numbers will have only two square roots, but this should be assume-able (and expected to be known) knowledge for 4U students.)
Last edited by InteGrand; 26 Feb 2017 at 1:34 PM.
I don't see why a longer working out method is more legitimate. In fact in mathematics, we are always looking for shorter and more elegant solutions.
In some questions where finding the square root of a complex number is but a small part of the overall problem, you will not want to go thru the rigmarole of the long-winded method.
Last edited by Drongoski; 26 Feb 2017 at 8:00 PM.
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