How do you solve # sqrtz=-1#?
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Square both sides, #sqrt(z)=-1#, #[(sqrt(z))^2=(-1)^2]-=[z=1]#
By definition, for #x# and #y# real numbers, #sqrt# denotes the principle square root. That is:
For #x >=0#,
#y = sqrtx# if and only if (1) #y^2 = x# AND (2) #y >= 0#.
It is true that every positive number has two square roots.
That means that for every positive number #n#, the equation #y^2 = n# has two solutions.
The square root symbol (the square root function) denotes the non-negative solution.
Although it is true that #1^2 = 1# and #(-1)^2 = 1#, the principle root of #1# is #1# (not #-1#).
There is no solution to #sqrtz = -1#
If we are working in the complex numbers , the square root function is unchanged for positive real numbers.
For negative real number, #z#, the notation #sqrtz# denotes the principle square root which is a complex number with positive imaginary part.
#sqrt(-n) = bi# if and only if
(1) #(bi)^2 = -n# , and
(2) #b > 0#
#sqrt1 = 1# #" "# (Not #-1#)
#sqrt4 = 2# #" "# (Not #-2#)
#sqrt(-1) = i# #" "# (Not #-i# which has imaginary part #-1#)
#sqrt(-9) = 3i# #" "# (Not #-3i# which has imaginary part #-3#)
For complex numbers more generally, there is no consensus on a "principle" square root.
So there is no principle square root of #3+4i# for instance. There are two square roots, but neither is "principle".
