How do you solve # sqrtz=-1#?

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Answer 1 · 2017-08-04T15:44:34

#z=1#

Square both sides, #sqrt(z)=-1#, #[(sqrt(z))^2=(-1)^2]-=[z=1]#

Answer 2 · 2017-08-04T16:02:48

There is no solution.

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".