複素数

2乗のルート

by nomura · 2021年9月9日

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2乗のルート

a \in R

とする。

(1)

\sqrt{α^{2}} = | α | \sqrt{{sgn}^{2} (α)}

(2)

\sqrt{a^{2}} = | a |

(3)

\sqrt{- a^{2}} = | a | i

(4)

\sqrt{\pm a^{2}} = \frac{1}{2} (1 \pm 1 + (1 \mp 1) i) | a |

(1)

\begin{aligned} \sqrt{α^{2}} & = \sqrt{{| α |}^{2} {sgn}^{2} (α)} \\ = | α | \sqrt{{sgn}^{2} (α)} \end{aligned}

(1)-2

\begin{aligned} \sqrt{α^{2}} & = e^{\frac{1}{2} Log (α^{2})} \\ = e^{\frac{1}{2} (\ln | α^{2} | + i Arg (α^{2}))} \\ = e^{\ln | α | + \frac{i}{2} Arg ({| α |}^{2} {sgn}^{2} (α))} \\ = | α | e^{\frac{i}{2} Arg ({sgn}^{2} (α))} \\ = | α | e^{\frac{1}{2} Log (sgn ({sgn}^{2} (α)))} \\ = | α | e^{\frac{1}{2} Log ({sgn}^{2} (α))} \\ = | α | \sqrt{{sgn}^{2} (α)} \end{aligned}

(2)

\begin{aligned} \sqrt{a^{2}} & = | a | \sqrt{{sgn}^{2} (a)} \\ = | a | \end{aligned}

(2)-2

\begin{aligned} \sqrt{a^{2}} & = e^{\frac{1}{2} Log (a^{2})} \\ = e^{\frac{1}{2} \ln | a^{2} |} \\ = e^{\frac{1}{2} \ln {| a |}^{2}} \\ = e^{\ln (| a |)} \\ = | a | \end{aligned}

(3)

\begin{aligned} \sqrt{- a^{2}} & = \sqrt{{(a i)}^{2}} \\ = | a i | \sqrt{{sgn}^{2} (a i)} \\ = | a | \sqrt{{sgn}^{2} (i)} \\ = | a | i \end{aligned}

(3)-2

\begin{aligned} \sqrt{- a^{2}} & = e^{\frac{1}{2} Log (- a^{2})} \\ = e^{\frac{1}{2} (\ln | - a^{2} | + i Arg (- a^{2}))} \\ = e^{\frac{1}{2} (\ln (a^{2}) + i π)} \\ = i e^{\frac{1}{2} (\ln ({| a |}^{2}))} \\ = i e^{\ln (| a |)} \\ = i | a | \end{aligned}

(4)

\begin{aligned} \sqrt{\pm a^{2}} & = \sqrt{e^{\frac{π i}{2} (1 \mp 1)} a^{2}} \\ = \sqrt{{(e^{\frac{π i}{4} (1 \mp 1)} a)}^{2}} \\ = | e^{\frac{π i}{4} (1 \mp 1)} a | \sqrt{{sgn}^{2} (e^{\frac{π i}{4} (1 \mp 1)} a)} \\ = | a | \sqrt{e^{\frac{π i}{2} (1 \mp 1)}} \\ = | a | e^{\frac{π i}{4} (1 \mp 1)} \\ = (\frac{1 + i}{\sqrt{2}}) (\frac{1 \mp i}{\sqrt{2}}) | a | \\ = \frac{1}{2} (1 + (1 \mp 1) i \mp (- 1)) | a | \\ = \frac{1}{2} (1 \pm 1 + (1 \mp 1) i) | a | \end{aligned}

(4)-2

\begin{aligned} \sqrt{\pm a^{2}} & = e^{\frac{1}{2} Log (\pm a^{2})} \\ = e^{\frac{1}{2} (Log (a^{2}) + Log (\pm 1))} \\ = e^{\frac{1}{2} (Log (a^{2}) + \frac{1 \mp 1}{2} i π)} \\ = e^{\frac{i π}{4}} e^{\frac{\mp i π}{4}} e^{\frac{1}{2} (Log ({| a |}^{2}))} \\ = e^{\frac{i π}{4} (1 \mp 1)} | a | \\ = \frac{1}{2} (1 \pm 1 + (1 \mp 1) i) | a | \end{aligned}

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冪乗の対数

Log α^{β} = ℜ (β) \ln | α | - ℑ (β) Arg (α) + \mod (ℜ (β) Arg (α) + ℑ (β) \ln | α |, - 2 π, π)

指数関数の実部と虚部

| α^{β} | = {| α |}^{ℜ (β)} e^{- ℑ (β) Arg (α)}

偏角・対数と符号関数の関係

Arg (z) = - i Log (sgn (z))

冪乗の性質

pv α^{β} pv α^{γ} = pv α^{β + γ}