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authorJarrod Millman <millman@berkeley.edu>2008-09-02 20:32:38 +0000
committerJarrod Millman <millman@berkeley.edu>2008-09-02 20:32:38 +0000
commitd29107a248130bbb68ecac58d6720d716529141c (patch)
tree9e1408025c80d63fe9fa6151343a7b265ec3f654 /numpy/lib/scimath.py
parenta4a7966cf7191c28a7054127fd3717d6c760d556 (diff)
downloadnumpy-d29107a248130bbb68ecac58d6720d716529141c.tar.gz
reindenting prior to release
Diffstat (limited to 'numpy/lib/scimath.py')
-rw-r--r--numpy/lib/scimath.py6
1 files changed, 3 insertions, 3 deletions
diff --git a/numpy/lib/scimath.py b/numpy/lib/scimath.py
index ed04d1ce5..0d765fa20 100644
--- a/numpy/lib/scimath.py
+++ b/numpy/lib/scimath.py
@@ -198,15 +198,15 @@ def sqrt(x):
As the numpy.sqrt, this returns the principal square root of x, which is
what most people mean when they use square root; the principal square root
- of x is not any number z such as z^2 = x.
+ of x is not any number z such as z^2 = x.
For positive numbers, the principal square root is defined as the positive
- number z such as z^2 = x.
+ number z such as z^2 = x.
The principal square root of -1 is i, the principal square root of any
negative number -x is defined a i * sqrt(x). For any non zero complex
number, it is defined by using the following branch cut: x = r e^(i t) with
- r > 0 and -pi < t <= pi. The principal square root is then
+ r > 0 and -pi < t <= pi. The principal square root is then
sqrt(r) e^(i t/2).
"""
x = _fix_real_lt_zero(x)