calculate 1)cos(1+i) , sin(1+3i) 2) arctan(i), arctan(2i) , arctan(1+i) ,arctan(1−i) , arctan(1+2i). 3) have us conj(arctanz)=arctan(z^− )?


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Question Number 73411 by mathmax by abdo last updated on 11/Nov/19

$c a l c u l a t e$ $1) c o s (1 + i), s i n (1 + 3 i)$ $2) a r c t a n (i), a r c t a n (2 i), a r c t a n (1 + i), a r c t a n (1 - i),$ $a r c t a n (1 + 2 i) .$ $3) h a v e u s c o n j (a r c t a n z) = a r c t a n (\bar{z}) ?$
Commented by mathmax by abdo last updated on 12/Nov/19
$1) we use formulas cos(z)=ch(iz) and sinz =sh(iz) ⇒ cos(1+i) =ch(i(1+i)) =ch(i−1) =((e^(i−1) +e^(−i+1) )/2) =((e^(−1) {cos1 +isin1} +e{cos1−isin1})/2) =(((e+e^(−1) )cos1 −i(e−e^(−1) )sin1)/2) =ch(1)cos1−ish(1)sin(1) sin(1+3i) =sh(i−3) =((e^(−3+i) −e^(3−i) )/2) =((e^(−3) {cos1 +isin1}−e^3 {cos1−isin1})/2) =(((e^(−3) −e^3 )cos(1) +i(e^(−3) +e^3 )sin1)/2) =−sh(3)cos1 +i ch(3)sin1$
$1) w e u s e f o r m u l a s c o s (z) = c h (i z) a n d s i n z = s h (i z) \Rightarrow$ $c o s (1 + i) = c h (i (1 + i)) = c h (i - 1) = \frac{e^{i - 1} + e^{- i + 1}}{2}$ $= \frac{e^{- 1} {c o s 1 + i s i n 1} + e {c o s 1 - i s i n 1}}{2}$ $= \frac{(e + e^{- 1}) c o s 1 - i (e - e^{- 1}) s i n 1}{2} = c h (1) c o s 1 - i s h (1) s i n (1)$ $s i n (1 + 3 i) = s h (i - 3) = \frac{e^{- 3 + i} - e^{3 - i}}{2}$ $= \frac{e^{- 3} {c o s 1 + i s i n 1} - e^{3} {c o s 1 - i s i n 1}}{2}$ $= \frac{(e^{- 3} - e^{3}) c o s (1) + i (e^{- 3} + e^{3}) s i n 1}{2}$ $= - s h (3) c o s 1 + i c h (3) s i n 1$
Commented by mathmax by abdo last updated on 12/Nov/19
$2) we have arctan(z)=(1/(2i))ln(((1+iz)/(1−iz))) ⇒ arctan(i)can not calculed by this formula arctan(2i) =(1/(2i))ln(((1+i(2i))/(1−i(2i))))=(1/(2i))ln(((−1)/3)) =(1/(2i))ln(−1)+(1/(2i))ln((1/3)) =(1/(2i))(iπ)−(1/(2i))ln(3) =(π/2) +(i/2)ln(3) arctan(1+i) =(1/(2i))ln(((1+i(1+i))/(1−i(1+i)))) =(1/(2i))ln(((1+i−1)/(1−i +1))) =(1/(2i))ln((i/(2−i))) =(1/(2i)){ln(i)−ln(2−i)} =(1/(2i)){ ((iπ)/2)−ln((√5)e^(−i arctan((1/2))) } =(π/4) −(1/(2i))(ln((√5)) −i arctan((1/2))) =(π/4) +(i/4)ln(5) +(1/2) arctan((1/2))=(π/4) +(1/2) arctan((1/2))+((ln(5))/4)i arctan(1−i) =(1/(2i))ln(((1+i(1−i))/(1−i(1−i)))) =(1/(2i))ln(((1+i+1)/(1−i−1))) =(1/(2i))ln(((2+i)/(−i))) =(1/(2i)){ln(2+i)−ln(−i)} =(1/(2i)){ ln((√5)e^(iarctan((1/2))) +(((iπ)/2))} =(1/(2i))ln((√5))+(1/(2i))(iarctan((1/2))) +(π/4) =−(i/4)ln(5) +(1/2) arctan((1/2))+(π/4) =(π/4) +(1/2) arctan((1/2))−((ln(5))/4)i we see that conj(arctan(1+i))=arctan(conj(1+i)) rest to prove this result in general case. arctan(1+2i) =(1/(2i))ln(((1+i(1+2i))/(1−i(1+2i)))) =(1/(2i))ln( ((1+i−2)/(1−i+2))) =(1/(2i))ln(((−1+i)/(3−i))) =(1/(2i)){ ln(−1+i)−ln(3−i)} =(1/(2i)){ ln((√2)e^(−iarctan(1)) )−ln((√(10))e^(−i arctan((1/3))) )} =(1/(2i)){ ln((√2))−i(π/4)−ln((√(10))) +i arctan((1/3))} =−(i/4)ln(2)+(i/4)ln(10) −(π/8) +(1/2) arctan((1/3)) =(i/4)ln(5) +(1/2)arctan((1/3))−((ln(2))/4) i don t forget the formula arctanz =(1/(2i))ln(((1+iz)/(1−iz))) z from C$
$2) w e h a v e a r c t a n (z) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) \Rightarrow a r c t a n (i) c a n n o t$ $c a l c u l e d b y t h i s f o r m u l a$ $a r c t a n (2 i) = \frac{1}{2 i} l n (\frac{1 + i (2 i)}{1 - i (2 i)}) = \frac{1}{2 i} l n (\frac{- 1}{3})$ $= \frac{1}{2 i} l n (- 1) + \frac{1}{2 i} l n (\frac{1}{3}) = \frac{1}{2 i} (i π) - \frac{1}{2 i} l n (3)$ $= \frac{π}{2} + \frac{i}{2} l n (3)$ $a r c t a n (1 + i) = \frac{1}{2 i} l n (\frac{1 + i (1 + i)}{1 - i (1 + i)}) = \frac{1}{2 i} l n (\frac{1 + i - 1}{1 - i + 1})$ $= \frac{1}{2 i} l n (\frac{i}{2 - i}) = \frac{1}{2 i} {l n (i) - l n (2 - i)}$ $= \frac{1}{2 i} {\frac{i π}{2} - l n (\sqrt{5} e^{- i a r c t a n (\frac{1}{2})}}$ $= \frac{π}{4} - \frac{1}{2 i} (l n (\sqrt{5}) - i a r c t a n (\frac{1}{2}))$ $= \frac{π}{4} + \frac{i}{4} l n (5) + \frac{1}{2} a r c t a n (\frac{1}{2}) = \frac{π}{4} + \frac{1}{2} a r c t a n (\frac{1}{2}) + \frac{l n (5)}{4} i$ $a r c t a n (1 - i) = \frac{1}{2 i} l n (\frac{1 + i (1 - i)}{1 - i (1 - i)}) = \frac{1}{2 i} l n (\frac{1 + i + 1}{1 - i - 1})$ $= \frac{1}{2 i} l n (\frac{2 + i}{- i}) = \frac{1}{2 i} {l n (2 + i) - l n (- i)}$ $= \frac{1}{2 i} {l n (\sqrt{5} e^{i a r c t a n (\frac{1}{2})} + (\frac{i π}{2})} = \frac{1}{2 i} l n (\sqrt{5}) + \frac{1}{2 i} (i a r c t a n (\frac{1}{2}))$ $+ \frac{π}{4} = - \frac{i}{4} l n (5) + \frac{1}{2} a r c t a n (\frac{1}{2}) + \frac{π}{4}$ $= \frac{π}{4} + \frac{1}{2} a r c t a n (\frac{1}{2}) - \frac{l n (5)}{4} i w e s e e t h a t$ $c o n j (a r c t a n (1 + i)) = a r c t a n (c o n j (1 + i)) r e s t t o p r o v e t h i s r e s u l t$ $i n g e n e r a l c a s e .$ $a r c t a n (1 + 2 i) = \frac{1}{2 i} l n (\frac{1 + i (1 + 2 i)}{1 - i (1 + 2 i)})$ $= \frac{1}{2 i} l n (\frac{1 + i - 2}{1 - i + 2}) = \frac{1}{2 i} l n (\frac{- 1 + i}{3 - i})$ $= \frac{1}{2 i} {l n (- 1 + i) - l n (3 - i)}$ $= \frac{1}{2 i} {l n (\sqrt{2} e^{- i a r c t a n (1)}) - l n (\sqrt{10} e^{- i a r c t a n (\frac{1}{3})})}$ $= \frac{1}{2 i} {l n (\sqrt{2}) - i \frac{π}{4} - l n (\sqrt{10}) + i a r c t a n (\frac{1}{3})}$ $= - \frac{i}{4} l n (2) + \frac{i}{4} l n (10) - \frac{π}{8} + \frac{1}{2} a r c t a n (\frac{1}{3})$ $= \frac{i}{4} l n (5) + \frac{1}{2} a r c t a n (\frac{1}{3}) - \frac{l n (2)}{4} i$ $d o n t f o r g e t t h e f o r m u l a$ $a r c t a n z = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) z f r o m C$
	Answered by MJS last updated on 11/Nov/19

	$formulas :$ $\sin (a + b i) = \sin a \cosh b + i \cos a \sinh b$ $\cos (a + b i) = \cos a \cosh b - i \sin a \sinh b$ $\tan (a + b i) = \frac{{2 e}^{2 b} \sin 2 a}{e^{4 b} + {2 e}^{2 b} \cos 2 a + 1} + \frac{e^{4 b} - 1}{e^{4 b} + {2 e}^{2 b} \cos 2 a + 1} i$ $\csc (a + b i) = \frac{{2 e}^{b} (e^{2 b} + 1) \sin a}{e^{4 b} - {2 e}^{2 b} \cos 2 a + 1} - \frac{{2 e}^{b} (e^{2 b} - 1) \cos a}{e^{4 b} - {2 e}^{2 b} \cos 2 a + 1} i$ $\sec (a + b i) = \frac{{2 e}^{b} (e^{2 b} + 1) \cos a}{e^{4 b} + {2 e}^{2 b} \cos 2 a + 1} + \frac{{2 e}^{b} (e^{2 b} - 1) \sin a}{e^{4 b} + {2 e}^{2 b} \cos 2 a + 1} i$ $\cot (a + b i) = \frac{{2 e}^{2 b} (e^{2 b} \sin 4 a + (e^{4 b} + 1) \sin 2 a)}{e^{8 b} - {2 e}^{4 b} \cos 4 a + 1} - \frac{(e^{4 b} - 1) (e^{4 b} + {2 e}^{2 b} \cos 2 a + 1)}{e^{8 b} - {2 e}^{4 b} \cos 4 a + 1}$
	Answered by MJS last updated on 12/Nov/19

	$formulas for the arc - functions are very$ $complicated but$ $\arctan (a + b i) = \frac{π}{2} sign (a) + \frac{1}{2} (\arctan \frac{b - 1}{a} - \arctan \frac{b + 1}{a}) + \frac{i}{4} \ln \frac{a^{2} + {(b + 1)}^{2}}{a^{2} + {(b - 1)}^{2}}$
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