is this right when (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) ? I had let arg(a+bi)= { ((tan^(−1) ((b/a))),(a≥0 and b≥0)),((π−tan^(−1) (−(b/a))),(a<0 and b≥0)),((−(π−tan^(−1) ((b/a)))),(a<0 and b<0)),((−tan^(−1) ((b/a))),(a≥0 and b<0)) :} before I solved it (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) =∣a+bi∣^c ∣a+bi∣^di e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi)) =∣a+bi∣^c (c^di )^(ln∣a+bi∣) e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi))


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 217101 by MathematicalUser2357 last updated on 01/Mar/25
$is this right when (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) ? I had let arg(a+bi)= { ((tan^(−1) ((b/a))),(a≥0 and b≥0)),((π−tan^(−1) (−(b/a))),(a<0 and b≥0)),((−(π−tan^(−1) ((b/a)))),(a<0 and b<0)),((−tan^(−1) ((b/a))),(a≥0 and b<0)) :} before I solved it (a+bi)^(c+di) =∣a+bi∣^(c+di) e^(i(c+di)arg(a+bi)) =∣a+bi∣^c ∣a+bi∣^di e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi)) =∣a+bi∣^c (c^di )^(ln∣a+bi∣) e^(ic∙arg(a+bi)) e^(−d∙arg(a+bi))$
$is this right when {(a + b i)}^{c + d i} =∣ a + b i ∣^{c + d i} e^{i (c + d i) \arg (a + b i)} ?$ $I had let \arg (a + b i) = {\begin{cases} \tan^{- 1} (\frac{b}{a}) & a ⩾ 0 and b ⩾ 0 \\ π - \tan^{- 1} (- \frac{b}{a}) & a < 0 and b ⩾ 0 \\ - (π - \tan^{- 1} (\frac{b}{a})) & a < 0 and b < 0 \\ - \tan^{- 1} (\frac{b}{a}) & a ⩾ 0 and b < 0 \end{cases} before I solved it$ ${(a + b i)}^{c + d i} =∣ a + b i ∣^{c + d i} e^{i (c + d i) \arg (a + b i)}$ $=∣ a + b i ∣^{c} ∣ a + b i ∣^{d i} e^{i c \cdot \arg (a + b i)} e^{- d \cdot \arg (a + b i)}$ $=∣ a + b i ∣^{c} {(c^{d i})}^{\ln ∣ a + b i ∣} e^{i c \cdot \arg (a + b i)} e^{- d \cdot \arg (a + b i)}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum