find-the-Re-w-and-Im-w-where-w-sin-a-icos-a-cos-a-isin-a-

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 (\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 \dots

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

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