find all solutions for z∈C z^i =(1/2)−(1/2)i z^(1−i) =1+i


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Question Number 60039 by MJS last updated on 17/May/19

$find all solutions for z \in C$ $z^{i} = \frac{1}{2} - \frac{1}{2} i$ $z^{1 - i} = 1 + i$
Commented by maxmathsup by imad last updated on 18/May/19
$we have z^i =e^(iln(z)) so if z=r e^(iθ) we get ln(z) =ln(r)+iθ ⇒z^i =e^(i{ln(r)+iθ}) =e^(−θ) e^(iln(r)) =e^(−θ) {cos(ln(r) +isin(ln(r)} z^i =(1/2) −(1/2)i ⇒ e^(−θ) cos(ln(r))=(1/2) and e^(−θ) sin(ln(r))=−(1/2) ⇒ e^(−2θ) {cos^2 (lnr) +sin^2 (ln(r))} =(1/4) +(1/4) =(1/2) ⇒e^(−2θ) =(1/2) ⇒−2θ =−ln(2) ⇒θ =((ln(2))/2) also we have ((sin(lnr))/(cos(lnr))) =−1 ⇒tan(lnr) =−1 ⇒ ln(r) =arctan(−1) =−(π/4) ⇒r =e^(−(π/4)) ⇒z =e^(−(π/4)) e^(i((ln(2))/2)) .$
$w e h a v e z^{i} = e^{i l n (z)} s o i f z = r e^{i θ} w e g e t$ $l n (z) = l n (r) + i θ \Rightarrow z^{i} = e^{i {l n (r) + i θ}} = e^{- θ} e^{i l n (r)} = e^{- θ} {c o s (l n (r) + i s i n (l n (r)}$ $z^{i} = \frac{1}{2} - \frac{1}{2} i \Rightarrow e^{- θ} c o s (l n (r)) = \frac{1}{2} a n d e^{- θ} s i n (l n (r)) = - \frac{1}{2} \Rightarrow$ $e^{- 2 θ} {{c o s}^{2} (l n r) + {s i n}^{2} (l n (r))} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \Rightarrow e^{- 2 θ} = \frac{1}{2} \Rightarrow - 2 θ = - l n (2)$ $\Rightarrow θ = \frac{l n (2)}{2} a l s o w e h a v e \frac{s i n (l n r)}{c o s (l n r)} = - 1 \Rightarrow t a n (l n r) = - 1 \Rightarrow$ $l n (r) = a r c t a n (- 1) = - \frac{π}{4} \Rightarrow r = e^{- \frac{π}{4}} \Rightarrow z = e^{- \frac{π}{4}} e^{i \frac{l n (2)}{2}} .$
Commented by maxmathsup by imad last updated on 18/May/19

$i n m y s o l u t i o n i h a v e t a k e n t h e p r i n c i p a l d e t e r m i n a t i o n o f l o g .$
Commented by maxmathsup by imad last updated on 19/May/19

$a l l s o l u t i o n f o r t h i s e q u a t i o n a r e Z_{n} = e^{- \frac{π}{4}} e^{i (\frac{l n (2)}{2} + 2 n π)}$
	Answered by tanmay last updated on 17/May/19
	$(A+iB)^(P+iQ) =e^((P+iQ)Log(A+iB)) Log(A+iB) A=rcosθ B=rsinθ Log(re^(iθ) ) =Log_e (re^(i(2kπ+θ)) =log_e r+i(2kπ+θ) =(1/2)log_e (A^2 +B^2 )+i[2kπ+tan^(−1) ((B/A))] e^((P+iQ)[(1/2)log(A^2 +B^2 )+i{2kπ+tan^(−1) ((B/A))}]) e^p ×e^([{−Q(2kπ+tan^(−1) ((B/A))}+i(Q/2)log(A^2 +B^2 )]) e^P ×e^(−Q(2kπ+tan^(−1) ((B/A))) ×e^(i×(Q/2)log(A^2 +B^2 )) e^(P−Q(2kπ+tan^(−1) ((B/A))) ×[cos(((Qlog(A^2 +B^2 ))/2))+isin(((Qlog(A^2 +B^2 ))/2))] wait sir... now put P=0 and Q=1 e^(−(2kπ+tan^(−1) ((B/A)))) ×[cos(((log(A^2 +B^2 ))/2))+isin(((log(A^2 +B^2 )/2))] still[to go...$
	${(A + i B)}^{P + i Q}$ $= e^{(P + i Q) L o g (A + i B)}$ $L o g (A + i B)$ $A = r c o s θ B = r s i n θ$ $L o g ({r e}^{i θ})$ $= {L o g}_{e} ({r e}^{i (2 k π + θ)}$ $= {l o g}_{e} r + i (2 k π + θ)$ $= \frac{1}{2} {l o g}_{e} (A^{2} + B^{2}) + i [2 k π + {t a n}^{- 1} (\frac{B}{A})]$ $e^{(P + i Q) [\frac{1}{2} l o g (A^{2} + B^{2}) + i {2 k π + {t a n}^{- 1} (\frac{B}{A})}]}$ $e^{p} \times e^{[{- Q (2 k π + {t a n}^{- 1} (\frac{B}{A})} + i \frac{Q}{2} l o g (A^{2} + B^{2})]}$ $e^{P} \times e^{- Q (2 k π + {t a n}^{- 1} (\frac{B}{A})} \times e^{i \times \frac{Q}{2} l o g (A^{2} + B^{2})}$ $e^{P - Q (2 k π + {t a n}^{- 1} (\frac{B}{A})} \times [c o s (\frac{Q l o g (A^{2} + B^{2})}{2}) + i s i n (\frac{Q l o g (A^{2} + B^{2})}{2})]$ $w a i t s i r . . .$ $n o w p u t P = 0 a n d Q = 1$ $e^{- (2 k π + {t a n}^{- 1} (\frac{B}{A}))} \times [c o s (\frac{l o g (A^{2} + B^{2})}{2}) + i s i n (\frac{l o g (A^{2} + B^{2}}{2})]$ $s t i l l [t o g o . . .$
	Commented by MJS last updated on 18/May/19

	$nice until here . thanks a lot!$ $I^{'} m really not sure how to solve these, will$ $look into it soon . . .$
	Answered by Smail last updated on 18/May/19

	$(1) z^{i} = \frac{1}{2} - \frac{1}{2} i = \frac{1}{\sqrt{2}} e^{i \frac{π}{4}} = 2^{- \frac{1}{2}} e^{i (\frac{π}{4} + 2 k π)}$ $z = 2^{- \frac{1}{2 i}} e^{\frac{π}{4} + 2 k π} = 2^{\frac{i}{2}} e^{\frac{π}{4} + 2 k π}$ $= e^{\frac{i}{2} l n 2} \times e^{\frac{π}{4} + 2 k π}$ $z = e^{\frac{π}{4} + 2 k π} e^{i \frac{l n 2}{2}} = e^{\frac{π}{4} + 2 k π} 2^{i / 2}$ $(2) z^{1 - i} = 1 + i = \sqrt{2} e^{i (\frac{π}{4} + 2 k π)} = 2^{\frac{1}{2}} e^{i (\frac{π}{4} + 2 k π)}$ $z = 2^{\frac{1}{2 (1 - i)}} e^{\frac{i}{1 - i} (π / 4 + 2 k π)} = 2^{\frac{1 + i}{4}} \times e^{\frac{i - 1}{2} (\frac{π}{4} + 2 k π)}$ $= 2^{\frac{1}{4}} \times e^{\frac{i}{4} l n 2} \times e^{\frac{i}{2} (\frac{π}{4} + 2 k π)} \times e^{- \frac{1}{2} (\frac{π}{4} + 2 k π)}$ $z = \sqrt[4]{2} e^{- \frac{π / 4 + 2 k π}{2}} \times e^{i (\frac{π}{4} + 2 k π + \frac{l n 2}{4})}$
	Commented by MJS last updated on 18/May/19

	$thank you so far . but which values of k are$ $appropriate ?$
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