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Thread: complex no. question no. 6

  1. #1
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    complex no. question no. 6

    p302 q19
    given
    $\displaystyle \frac{1+sinx +icosx}{1+sinx -icosx} = sin x + i cosx $
    evaluate:
    $\displaystyle (1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5$

    my working
    $\displaystyle =(-i)((1+sin\frac\pi 5 +icos\frac\pi 5)^5+(1+sin\frac\pi 5 +icos\frac\pi 5)^5)$
    $\displaystyle =(-i)(2(1+sin\frac\pi 5)^5-20(1+sin\frac\pi 5)^3 cos^2 \frac\pi 5 +5(1+sin \frac\pi 5) cos^4 \frac\pi 5 )$
    and stuck.
    thanks!
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    p302 q19
    given
    $\displaystyle \frac{1+sinx +icosx}{1+sinx -icosx} = sin x + i cosx $
    evaluate:
    $\displaystyle (1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5$

    my working
    $\displaystyle =(-i)((1+sin\frac\pi 5 +icos\frac\pi 5)^5+(1+sin\frac\pi 5 +icos\frac\pi 5)^5)$
    $\displaystyle =(-i)(2(1+sin\frac\pi 5)^5-20(1+sin\frac\pi 5)^3 cos^2 \frac\pi 5 +5(1+sin \frac\pi 5) cos^4 \frac\pi 5 )$
    and stuck.
    thanks!
    Let :

    $\displaystyle N=(1+sin\frac\pi 5 +icos\frac\pi 5)^5+ i (1+sin\frac\pi 5 -icos\frac\pi 5)^5$

    Because you're given the formula above, divide by $\displaystyle (1+\sin \frac \pi 5-i \cos \frac \pi 5)^5$ (because it's different to 0, we can do it)

    Therefore :

    $\displaystyle \begin{aligned} \frac{N}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5} &=\frac{(1+sin\frac\pi 5 +icos\frac\pi 5)^5}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5}+i \frac{(1+sin\frac\pi 5 -icos\frac\pi 5)^5}{(1+sin\frac\pi 5 -icos\frac\pi 5)^5} \\
    &=\left(\frac{1+sin\frac\pi 5 +icos\frac\pi 5}{1+\sin \frac \pi 5-i \cos \frac \pi 5}\right)^5+i \\
    &=(\sin \frac \pi 5+i \cos \frac \pi 5)^5+i \end{aligned}$


    Now, it would be great if it could be simplified.
    You can notice that it looks like a lot de Moivre's formula :

    $\displaystyle (\cos x+i \sin x)^n=\cos nx+i \sin nx$

    So we're gonna make a transform... According to the unit circle (or to your lesson) :

    $\displaystyle \cos (\frac \pi 2-x)=\sin x$

    $\displaystyle \sin (\frac \pi 2-x)=\cos x$

    Great !

    Therefore $\displaystyle \sin \frac \pi 5+i \cos \frac \pi 5=\cos (\frac \pi 2-\frac \pi 5)+i \sin (\frac \pi 2-\frac \pi 5)=\cos \frac{3 \pi}{10}+i \sin \frac{3 \pi}{10}$

    Now we can apply de Moivre's formula :

    $\displaystyle \begin{aligned} (\sin \frac \pi 5+i \cos \frac \pi 5)^5 &=\left(\cos \frac{3 \pi}{10}+i \sin \frac{3 \pi}{10}\right)^5 \\ \\
    &=\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2} \\
    &=0+i (-1) \\
    &=-i \end{aligned}$

    -------------------

    $\displaystyle \implies \boxed{\frac{N}{(1+\sin \frac \pi 5-i \cos \frac \pi 5)^5}=-i+i}$


    Conclude
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  3. #3
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    thanks again!
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