find the Re(w) and Im(w) where w = (sin a + icos a)^((cos a + isin a))


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Question Number 73918 by arkanmath7@gmail.com last updated on 16/Nov/19

$f i n d t h e R e (w) a n d I m (w)$ $w h e r e w = {(s i n a + i c o s a)}^{(c o s a + i s i n a)}$
Commented by abdomathmax last updated on 16/Nov/19
$W=(e^(i((π/2)−a)) )^(cosa +isina) =e^(i((π/2)−a)(cosa +isina)) =e^(i((π/2)−a)cosa−((π/2)−a)sina) =e^((a−(π/2))sina) ×e^(i((π/2)−a)cosa) =e^((a−(π/2))sina) × {cos{((π/2)−a)cosa}+isin{((π/2)−a)cosa} ⇒Re(W)=e^((a−(π/2))sina) . cos{((π/2)−a)cosa} Im(W)=e^((a−(π/2))sina) .sin{((π/2)−a)cosa}$
$W = {(e^{i (\frac{π}{2} - a)})}^{c o s a + i s i n a} = e^{i (\frac{π}{2} - a) (c o s a + i s i n a)}$ $= e^{i (\frac{π}{2} - a) c o s a - (\frac{π}{2} - a) s i n a}$ $= e^{(a - \frac{π}{2}) s i n a} \times e^{i (\frac{π}{2} - a) c o s a}$ $= e^{(a - \frac{π}{2}) s i n a} \times {c o s {(\frac{π}{2} - a) c o s a} + i s i n {(\frac{π}{2} - a) c o s a}$ $\Rightarrow R e (W) = e^{(a - \frac{π}{2}) s i n a} . c o s {(\frac{π}{2} - a) c o s a}$ $I m (W) = e^{(a - \frac{π}{2}) s i n a} . s i n {(\frac{π}{2} - a) c o s a}$
Commented by arkanmath7@gmail.com last updated on 18/Nov/19

$t h n x s i r$
	Answered by Tanmay chaudhury last updated on 16/Nov/19

	${(A + i B)}^{P + i Q} = e^{(P + i Q) L o g (A + i B)}$ $l e t A = r c o s θ B = r s i n θ s o A + i B = {r e}^{i θ} = e^{i (2 n π + θ)}$ $r = \sqrt{A^{2} + B^{2}} t a n θ = \frac{B}{A} ⇛ θ = {t a n}^{-} (\frac{B}{A})$ $e^{(P + i Q) L o g ({r e}^{i (2 n π + θ)})}$ $= e^{(P + i Q) [l o g \sqrt{A^{2} + B^{2}} + i (2 n π + {t a n}^{- 1} (\frac{B}{A})]}$ $e^{[P . l o g \sqrt{A^{2} + B^{2}} + i (2 n π P + {P t a n}^{- 1} (\frac{B}{A}) + l o g \sqrt{A^{2} + B^{2}}) - Q (2 n π + {t a n}^{- 1} (\frac{B}{A}))}$ $= e^{[P . l o g \sqrt{A^{2} + B^{2}} - Q (2 n π + {t a n}^{- 1} (\frac{B}{A}))} \times e^{i (2 n π P + {P t a n}^{- 1} (\frac{B}{A}) + l o g \sqrt{A^{2} + B^{2}})}$ $n o w b a c k t o p r o b l e m$ $A = s i n a B = c o s a P = c o s a Q = s i n a$ $l o g \sqrt{A^{2} + B^{2}} = \frac{1}{2} l o g ({s i n}^{2} a + {c o s}^{2} a) = 0$ ${t a n}^{- 1} (\frac{B}{A}) = {t a n}^{- 1} (\frac{c o s a}{s i n a}) = {t a n}^{- 1} (t a n (\frac{π}{2} - a)) = \frac{π}{2} - a$ $s o p u t t i n g t h e v a l u e$ $= e^{[P . 0 - s i n a (2 n π + \frac{π}{2} -)]} \times e^{i (2 n π . c o s a + c o s a . (\frac{π}{2} - a)}$ $= e^{[- (2 n π + \frac{π}{2}) s i n a} \times [c o s (2 n π . c o s a + (\frac{π}{2} - a) c o s a) +$ $s i n (2 n π . c o s a + (\frac{π}{2} - a) c o s a]$ $p l s c h e c k u p t o t h i s . . .$
	Commented by arkanmath7@gmail.com last updated on 16/Nov/19

	$t h n x a l o t . {i t}^{'} s a l o n g s o l, u m s t b e t i r e d n o w$ $t h n x a g a i n$
	Commented by Tanmay chaudhury last updated on 16/Nov/19

	$n o i h a v e s o l v e d i n g e n e r a l w a y n o t d i r e c t l y$ ${(A + i B)}^{P + i Q} f o r m a t . .$ $i c a n s o l v e t h e g i v e n p r o b l e m d i r e c l y . .$ $b u t s o l v e d {(A + i B)}^{P + i Q} f o r m a t f o r h e l p i n o t h e r$ $p r o b l e m$
	Answered by Tanmay chaudhury last updated on 16/Nov/19
	$short method [cos((π/2)−a)+isin((π/2)−a)]^((cosa+isina)) =e^(i((π/2)−a)(cosa+isina)) =e^((a−(π/2))sina+i((π/2)−a)cosa) =e^((a−(π/2))sina) ×[cos{cosa((π/2)−a)}+isin{cosa((π/2)−a)}] =e^((a−(π/2))sina) cos{cosa((π/2)−a)}+i e^((a−(π/2))sina) sin{cosa((π/2)−a)}$
	$s h o r t m e t h o d$ ${[c o s (\frac{π}{2} - a) + i s i n (\frac{π}{2} - a)]}^{(c o s a + i s i n a)}$ $= e^{i (\frac{π}{2} - a) (c o s a + i s i n a)}$ $= e^{(a - \frac{π}{2}) s i n a + i (\frac{π}{2} - a) c o s a}$ $= e^{(a - \frac{π}{2}) s i n a} \times [c o s {c o s a (\frac{π}{2} - a)} + i s i n {c o s a (\frac{π}{2} - a)}]$ $= e^{(a - \frac{π}{2}) s i n a} c o s {c o s a (\frac{π}{2} - a)} + i e^{(a - \frac{π}{2}) s i n a} s i n {c o s a (\frac{π}{2} - a)}$
	Commented by peter frank last updated on 17/Nov/19

	$t h a n k s$
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