Editing Imaginary unit (section)

=== Roots ===
[[File:Imaginary2Root.svg|thumb|right|200px|The two square roots of {{mvar|i}} in the complex plane]]
Just like all nonzero complex numbers, <math display=inline>i = e^{\pi i/ 2}</math> has two distinct [[square root]]s which are [[additive inverse]]s. In polar form, they are
<math display=block>\begin{alignat}{3}
\sqrt{i} &= {\exp}\bigl(\tfrac12{\pi i}\bigr)^{1/2} &&{}= {\exp}\bigl(\tfrac14\pi i\bigr), \\
-\sqrt{i} &= {\exp}\bigl(\tfrac14{\pi i}-\pi i\bigr) &&{}= {\exp}\bigl({-\tfrac34\pi i}\bigr).
\end{alignat}</math>

In rectangular form, they are{{efn|To find such a number, one can solve the equation {{math|1=(''x'' + ''iy''){{sup|2}} = ''i''}} where {{mvar|x}} and {{mvar|y}} are real parameters to be determined, or equivalently {{math|1=''x''{{isup|2}} + 2''ixy'' − ''y''{{isup|2}} = ''i''.}} Because the real and imaginary parts are always separate, we regroup the terms, {{math|1=''x''{{isup|2}} − ''y''{{isup|2}} + 2''ixy'' = 0 + ''i''.}} By [[equating coefficients]], separating the real part and imaginary part, we have a system of two equations:

<math display=block>\begin{align}
x^{2} - y^{2} &= 0 \\[3mu]
2xy &= 1.
\end{align}</math>

Substituting <math display=inline>y=\tfrac12x^{-1}</math> into the first equation, we get <math display=inline> x^{2} - \tfrac14x^{-2} = 0</math> <math display=inline>\implies 4x^4 = 1.</math> Because {{mvar|x}} is a real number, this equation has two real solutions for {{mvar|x}}<math display=block>x=\tfrac{1}{\sqrt{2}}</math> and <math>x=-\tfrac{1}{\sqrt{2}}</math>. Substituting either of these results into the equation {{math|1=2''xy'' = 1}} in turn, we will get the corresponding result for {{mvar|y}}. Thus, the square roots of {{mvar|i}} are the numbers <math>\tfrac{1}{\sqrt{2}} + \tfrac{1}{\sqrt{2}}i</math> and <math>-\tfrac{1}{\sqrt{2}}-\tfrac{1}{\sqrt{2}}i</math>.<ref>{{cite web |website=University of Toronto Mathematics Network |title=What is the square root of {{mvar|i}}&nbsp;? |access-date=26 March 2007 |url=http://www.math.utoronto.ca/mathnet/questionCorner/rootofi.html}}</ref>}}

<math display=block>\begin{alignat}{3}
\sqrt{i} &= \frac{1 + i}{ \sqrt{2}} &&{}= \phantom{-}\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}i, \\[5mu]
-\sqrt{i} &= - \frac{1 + i}{ \sqrt{2}} &&{}= - \tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2}i.
\end{alignat}</math>

Squaring either expression yields
<math display=block>
\left( \pm \frac{1 + i}{\sqrt2} \right)^2
= \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i.
</math>

[[File:Imaginary3Root.svg|thumb|right|200px|The three cube roots of {{mvar|i}} in the complex plane]]
The three [[cube root]]s of {{mvar|i}} are<ref>{{Cite book |last1=Zill |first1=Dennis G. |url=https://www.worldcat.org/oclc/50495529 |title=A first course in complex analysis with applications |last2=Shanahan |first2=Patrick D. |year=2003 |publisher=Jones and Bartlett |isbn=0-7637-1437-2 |location=Boston |pages=24–25 |oclc=50495529}}</ref>

<math display=block>
\sqrt[3]i = {\exp}\bigl(\tfrac16 \pi i\bigr) = \tfrac{\sqrt{3}}{2} + \tfrac12i, \quad
            {\exp}\bigl(\tfrac56 \pi i\bigr) = -\tfrac{\sqrt{3}}{2} + \tfrac12i, \quad
            {\exp}\bigl({-\tfrac12 \pi i}\bigr) = -i.
</math>

For a general positive integer {{mvar|n}}, the [[nth root|{{mvar|n}}-th roots]] of {{mvar|i}} are, for {{math|1=''k'' = 0, 1, ..., ''n'' − 1,}}
<math display=block>
\exp \left(2 \pi i \frac{k+\frac14}{n} \right)
= \cos \left(\frac{4k+1}{2n}\pi \right) +
  i \sin \left(\frac{4k+1}{2n}\pi \right).
</math>
The value associated with {{math|1=''k'' = 0}} is the [[principal value|principal]] {{mvar|n}}-th root of {{mvar|i}}.  The set of roots equals the corresponding set of [[root of unity|roots of unity]] rotated by the principal {{mvar|n}}-th root of {{mvar|i}}. These are the vertices of a [[regular polygon]] inscribed within the complex [[unit circle]].