# Complex Numbers! (1 Viewer)

#### harshsyd

##### New Member
Hey, I got two questions;

I'll attach the images as I cannot use LaTeX for some reason.

1. https://imgur.com/a/CFbQErW (I tried expanding and simplifying using bionomial, I do get correct answers but of course they are not in the forms of cot and so on, I get straight square roots, expected).

2. https://imgur.com/a/eF2Su32 (The only real place I am struggling is the last part, deducing the cos(2O) statement, rest is completed).

Thank You!

#### Drongoski

##### Well-Known Member
Q5

$(z-1)^4 + (z+1)^4 = 0 \implies (z-1)^4 = -(z+1)^4 \implies \left ( \frac {z-1}{z+1} \right )^4 = -1 \\ \\ = cis(2k\pi + \pi) \implies \frac {z-1}{z+1} = cis(\frac {2k\pi + \pi}{4}) = cis\frac {\pi}{4}, cis \frac{3\pi}{4}, cis \frac {-\pi}{4} and cis \frac {-3\pi}{4}.\\ \\ now if \frac {z-1}{z+1} = cis \theta then z = \frac {1+cis \theta}{1 - cis \theta} = \frac {1 + cos\theta + isin\theta}{1-cos\theta - isin\theta} \\ \\ = \frac {2cos^2(\frac {\theta}{2}) + 2isin(\frac{\theta}{2})cos(\frac {\theta}{2})}{2sin^2(\frac {\theta}{2})cos(\frac {\theta}{2})} = \frac {2cos( \frac {\theta }{2})(cos(\frac {\theta}{2})+isin(\frac{\theta}{2}))}{2sin(\frac {\theta}{2})(sin(\frac{\theta}{2}) - icos(\frac {\theta}{2}))} = \cdots = icot(\frac {\theta}{2})\\ \\ \therefore z = icot(\frac{\pi}{8}), icot(\frac {3\pi}{8}),icot(\frac {-\pi}{8}) and icot(\frac {-3\pi}{8})\\ \\ = the desired result.$

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#### harshsyd

##### New Member
Thank you buddy, that was very nicely explained!
Except, how was the second last line derived? I understood up till 1+cosO+isinO/1-cosO-isinO...the half angles one, I am slightly lost

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#### fan96

##### 617 pages
$1 + \cos \theta = \left(\sin^2{\theta\over2} + \cos^2{\theta\over2}\right)+\left( \cos^2{\theta\over2} - \sin^2{\theta\over2}\right) = 2 \cos^2{\theta\over2}$

$1 - \cos \theta = \left(\sin^2{\theta\over2} + \cos^2{\theta\over2}\right)-\left( \cos^2{\theta\over2} - \sin^2{\theta\over2}\right) = 2 \sin^2{\theta\over2}$

I think the start of that second last line you were talking about should read:

$\frac{2 \cos^2 {\theta \over 2}+ i(2 \sin{\theta \over 2} \cos {\theta \over 2})}{2 \sin^2 {\theta \over 2} - i(2\sin{\theta \over 2}\cos{\theta \over 2})}$

#### fan96

##### 617 pages

Let $z = {\rm cis\,} \theta$.

${\rm cis \,} 4\theta + 1 = ({\rm cis\,} 2\theta - \sqrt2{\rm cis\,}\theta + 1)({\rm cis\,} 2\theta + \sqrt2{\rm cis\,}\theta + 1)$

Note:
\begin{aligned} ({\rm cis\,} 2\theta + \sqrt2{\rm cis\,}\theta + 1) &=[(\cos 2\theta + 1 + \sqrt 2 \cos \theta) + i(\sin 2\theta + \sqrt 2 \sin\theta) \\ &= (2 \cos^2 \theta + \sqrt 2 \cos \theta) + i(2\sin \theta \cos\theta+ \sqrt 2 \sin\theta)\\ &= 2\cos\theta( \cos \theta + {\sqrt2/2}) + i[2 \sin \theta(\cos\theta + {\sqrt2/2})]\end{aligned}

Similarly,
$({\rm cis\,} 2\theta - \sqrt2{\rm cis\,}\theta + 1) = 2\cos\theta( \cos \theta- {\sqrt2/2}) + i[2 \sin \theta(\cos\theta - {\sqrt2/2})]$

Going back to the original identity: multiplying the expression on the RHS and taking the real part of both sides,

\begin{aligned}\cos4\theta + 1 &= 4 \cos^2\theta( \cos^2 \theta - (\sqrt2/2)^2) - 4 \sin^2\theta( \cos^2 \theta - (\sqrt2/2)^2) \\ &= 4 \cos2\theta( \cos \theta + {\sqrt2/2}) ( \cos \theta - {\sqrt2/2}) \end{aligned}

And a few more lines of working give the required identity.

Thanks!