1. ## help solve (i-j)^n=(iz-1)^n

Hi,

I can't solve (i-j)^n=(iz-1)^n nor (z+i)^n + (z-i)^n =0.

I have tried dividing through and equating with the nth root but it hasn't gotten me anywhere?

Anoone know what to do??

Thanks..

2. $(i-z)^n=(iz-1)^n$

Let $z=x+yi$.

Apply modulus in both members:
$|i-z|=|iz-1]\Rightarrow|-x+(1-y)i|=|-1-y+xi|\Rightarrow$ $\Rightarrow x^2+1-2y+y^2=1+2y+y^2+x^2\Rightarrow y=0\Rightarrow z=x$

Divide equation by the right member:

$\left(\frac{i-x}{ix+1}\right)^n=1\Rightarrow\frac{i-x}{ix-1}=z_k$, where $z_k=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}, \ k\in{0,1,....,n-1}$

Find x: $x=\frac{i+z_k}{1+iz_k}$

$x=\frac{\cos\frac{2k\pi}{n}+i\left(1+\sin\frac{2k\ pi}{n}\right)}{1-\sin\frac{2k\pi}{n}+i\cos\frac{2k\pi}{n}}=$

$=\frac{\cos^2\frac{k\pi}{n}-\sin^2\frac{k\pi}{n}+i\left(\cos\frac{k\pi}{n}+\si n\frac{k\pi}{n}\right)^2}{\left(\cos\frac{k\pi}{n}-\sin\frac{k\pi}{n}\right)^2+i\left(\cos^2\frac{k\p i}{n}-\sin^2\frac{k\pi}{n}\right)}=$

$=\frac{\left(\cos\frac{k\pi}{n}+\sin\frac{k\pi}{n} \right)\left(\cos\frac{k\pi}{n}-\sin\frac{k\pi}{n}+i\left(\cos\frac{k\pi}{n}+\sin\ frac{k\pi}{n}\right)\right)}{\left(\cos\frac{k\pi} {n}-\sin\frac{k\pi}{n}\right)\left(\cos\frac{k\pi}{n}-\sin\frac{k\pi}{n}+i\left(\cos\frac{k\pi}{n}+\sin\ frac{k\pi}{n}\right)\right)}=$

$=\frac{\cos\frac{k\pi}{n}+\sin\frac{k\pi}{n}}{\cos \frac{k\pi}{n}-\sin\frac{k\pi}{n}}=\frac{1+\tan\frac{k\pi}{n}}{1-\tan\frac{k\pi}{n}}=\tan\left(\frac{\pi}{4}+\frac{ k\pi}{n}\right)$

Solve the other equation in a similar way.

3. big thanks

4. (tears of joy...) big thanks!