find-all-solutions-for-z-C-z-i-1-2-1-2-i-z-1-i-1-i-

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

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 + 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 \dots

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 \dots

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 \dots

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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