Complex number help (1 Viewer)

Run hard@thehsc · Nov 28, 2021

How would you approach and do 13b?

5uckerberg · Nov 28, 2021

Run hard@thehsc said:
View attachment 34304
How would you approach and do 13b?

Binomial theorem but with the complex number I.

matutor · Nov 28, 2021

Run hard@thehsc said:
View attachment 34304
How would you approach and do 13b?

Why not use de Moivre's theorem? Comes down to solving a simple trig equation.

Drongoski · Nov 28, 2021

a)
$r = \sqrt 2 \\ \\ (1+i)^n + (1-i)^n = (rcis\frac {\pi}{4})^n +(rcis{\frac{-\pi}{4}})^n \\ \\ = r^n cis \frac{n\pi}{4}) + r^n cis \frac{-n\pi}{4}\\ \\ =r^n(cos \frac {n\pi}{4} + isin \frac {n\pi}{4}) + r^n(cos \frac{n\pi}{4} - isin \frac{n\pi}{4})\\ \\ = 2r^n cos \frac {n\pi}{4}\\ \$
= real

b)
$\therefore 2 r^n cos \frac{n\pi}{4} = 0 \implies cos \frac {n\pi}{4} = 0 \implies \frac {n\pi}{4} = \pm (k\pi + \frac {\pi}{2})\\ \\ \implies n = \pm(4k + 2) = \pm(2, 6, 10, \cdots ) k = 0, 1, 2, 3, . . .$

CM_Tutor · Nov 29, 2021

There is a typo, FYI, in (b):

Drongoski said:
$\therefore 2 r^n cos \frac{n\pi}{4} = 0 \implies \color{blue} cos \frac {n\pi}{4} = 0 \implies \frac {n\pi}{4} = \pm (k\pi + \frac {1}{2}) \\ \\ \implies n = \pm(4k + 2) \color{black} = \pm(2, 6, 10, \cdots ) k = 0, 1, 2, 3, . . .$

The part in blue should be

$\cos{\cfrac{n\pi}{4}} = 0 \implies \cfrac{n\pi}{4} = \pm \left(k + \cfrac{1}{2}\right)\pi \implies n = \pm(4k + 2)$

Drongoski · Nov 29, 2021

CM_Tutor said:
There is a typo, FYI, in (b):

The part in blue should be

$\cos{\cfrac{n\pi}{4}} = 0 \implies \cfrac{n\pi}{4} = \pm \left(k + \cfrac{1}{2}\right)\pi \implies n = \pm(4k + 2)$

Thank you very much for pointing out my typo. Now corrected.

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Complex number help (1 Viewer)

Run hard@thehsc

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5uckerberg

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matutor

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Drongoski

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CM_Tutor

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Drongoski

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