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Question Number 216585 by MathematicalUser2357 last updated on 11/Feb/25
is this right? i had let θ= { ((tan^(−1) ((d/c))),((c>0,d>0))),((π−tan^(−1) ((d/c))),((c<0,d>0))),((−π+tan^(−1) ((d/c))),((c<0,d<0))),((−tan^(−1) ((d/c))),((c>0,d<0))) :} before i calculated below (a+bi)^(c+di) =∣a+di∣^(c+di) e^(i(c+di)θ) =∣a+bi∣^c ∣a+bi∣^di e^(icθ) e^(−dθ) =((√(a^2 +b^2 )))^c ((√(a^2 +b^2 )))^di e^(icθ) e^(−dθ) =((√(a^2 +b^2 )))^c e^((1/2)idln(a^2 +b^2 )) e^(icθ) e^(−dθ) =((√(a^2 +b^2 )))^c e^(−dθ) e^(i(dln(a^2 +b^2 )+cθ)) =((√(a^2 +b^2 )))^c e^(−dθ) (cos(dln(a^2 +b^2 )+cθ)+isin(dln(a^2 +b^2 )+cθ))
$$\mathrm{is}\:\mathrm{this}\:\mathrm{right}? \\ $$$$\mathrm{i}\:\mathrm{had}\:\mathrm{let}\:\theta=\begin{cases}{\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}>\mathrm{0},{d}>\mathrm{0}\right)}\\{\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}<\mathrm{0},{d}>\mathrm{0}\right)}\\{−\pi+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}<\mathrm{0},{d}<\mathrm{0}\right)}\\{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{d}}{{c}}\right)}&{\left({c}>\mathrm{0},{d}<\mathrm{0}\right)}\end{cases}\:\mathrm{before}\:\mathrm{i}\:\mathrm{calculated}\:\mathrm{below} \\ $$$$\left({a}+{bi}\right)^{{c}+{di}} =\mid{a}+{di}\mid^{{c}+{di}} {e}^{{i}\left({c}+{di}\right)\theta} \\ $$$$=\mid{a}+{bi}\mid^{{c}} \mid{a}+{bi}\mid^{{di}} {e}^{{ic}\theta} {e}^{−{d}\theta} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} \left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{di}} {e}^{{ic}\theta} {e}^{−{d}\theta} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} {e}^{\frac{\mathrm{1}}{\mathrm{2}}{id}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} {e}^{{ic}\theta} {e}^{−{d}\theta} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} {e}^{−{d}\theta} {e}^{{i}\left({d}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{c}\theta\right)} \\ $$$$=\left(\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{{c}} {e}^{−{d}\theta} \left(\mathrm{cos}\left({d}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{c}\theta\right)+{i}\mathrm{sin}\left({d}\mathrm{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{c}\theta\right)\right) \\ $$
Commented by Ghisom last updated on 12/Feb/25
x, y ∈C to calculate x^y you need these forms: x=re^(iθ) =e^(ln r +iθ) y=a+bi ⇒ x^y =(e^(ln r +iθ) )^(a+bi) =e^((ln r +iθ)(a+bi)) = =e^(aln r −bθ +i(bln r +aθ)) = =(r^a /e^(bθ) )(cos (aθ+bln r) +i sin (aθ+bln r)) your path is not wrong but too complicated
$${x},\:{y}\:\in\mathbb{C} \\ $$$$\mathrm{to}\:\mathrm{calculate}\:{x}^{{y}} \:\mathrm{you}\:\mathrm{need}\:\mathrm{these}\:\mathrm{forms}: \\ $$$${x}={r}\mathrm{e}^{\mathrm{i}\theta} =\mathrm{e}^{\mathrm{ln}\:{r}\:+\mathrm{i}\theta} \\ $$$${y}={a}+{b}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}^{{y}} =\left(\mathrm{e}^{\mathrm{ln}\:{r}\:+\mathrm{i}\theta} \right)^{{a}+{b}\mathrm{i}} =\mathrm{e}^{\left(\mathrm{ln}\:{r}\:+\mathrm{i}\theta\right)\left({a}+{b}\mathrm{i}\right)} = \\ $$$$=\mathrm{e}^{{a}\mathrm{ln}\:{r}\:−{b}\theta\:+\mathrm{i}\left({b}\mathrm{ln}\:{r}\:+{a}\theta\right)} = \\ $$$$=\frac{{r}^{{a}} }{\mathrm{e}^{{b}\theta} }\left(\mathrm{cos}\:\left({a}\theta+{b}\mathrm{ln}\:{r}\right)\:+\mathrm{i}\:\mathrm{sin}\:\left({a}\theta+{b}\mathrm{ln}\:{r}\right)\right) \\ $$$$\mathrm{your}\:\mathrm{path}\:\mathrm{is}\:\mathrm{not}\:\mathrm{wrong}\:\mathrm{but}\:\mathrm{too}\:\mathrm{complicated} \\ $$
Commented by MathematicalUser2357 last updated on 25/Feb/25
It′s time for part two
$$\mathrm{It}'\mathrm{s}\:\mathrm{time}\:\mathrm{for}\:\mathrm{part}\:\mathrm{two} \\ $$
Commented by MathematicalUser2357 last updated on 24/Feb/25

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