Prove the equation in complex numbers [on hold]

Which of the following equations is true in the complex numbers? Prove it.

(1+itana1−itana)n=1+itana1−itana\Bigl(\frac{1+i \tan a}{1-i \tan a}\Bigr)^n = \frac{1+i \tan a}{1-i \tan a}

(1+itana1−itana)n=1+itanna1−itanna\Bigl(\frac{1+i \tan a}{1-i \tan a}\Bigr)^n = \frac{1+ i \tan na}{1- i \tan na}

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3 Answers
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Note that
1+itana=1+isinacosa=cosa+isinacosa1−itana=1−isinacosa=cos(−a)+isin(−a)cosa\begin{align}
1+i\tan a&=1+i\frac{\sin a}{\cos a}=
\frac{\cos a+i\sin a}{\cos a}
\\[6px]
1-i\tan a&=1-i\frac{\sin a}{\cos a}=
\frac{\cos(-a)+i\sin(-a)}{\cos a}
\end{align}
Now recall De Moivre’s formula.

By the way, you can easily disprove the first equality by using a=π/4a=\pi/4 and n=2n=2, because in this case the left-hand side is
(1+i1−i)2≠1+i1−i
\Bigl(\frac{1+i}{1-i}\Bigr)^2\ne\frac{1+i}{1-i}

Of course the (second) equality holds provided tana\tan a and tanna\tan na exist.

Ok, first lets consider the term 1+itan(a)1−itan(a)\frac{1+i\tan(a)}{1-i\tan(a)}
if we multpily both top and bottom by 1+itan(a)1+i\tan(a) we get (1+itan(a))21+tan2(a)\frac{(1+i\tan(a))^2}{1 + \tan^2(a)}

and by using the identity sec2(a)=1+tan2(a)\sec^2(a) = 1 + \tan^2(a) and expanding the numerator we get
cos2(a)+2itan(a)cos2(a)−tan2(a)cos2(a)\cos^2(a)+2i\tan(a)\cos^2(a) – \tan^2(a)\cos^2(a)
=cos2(a)+2isin(a)cos(a)−sin2(a)= \cos^2(a) + 2i\sin(a)\cos(a) – \sin^2(a)
⟹1+itan(a)1−itan(a)=(cos(a)+isin(a))2\implies \frac{1+i\tan(a)}{1-i\tan(a)} = (\cos(a) + i\sin(a))^2

Ok now since we have this we can work out what (1+itan(a)1−itan(a))n(\frac{1+i\tan(a)}{1-i\tan(a)})^n is
(1+itan(a)1−itan(a))n=(cos(a)+(isin(a))2)n(\frac{1+i\tan(a)}{1-i\tan(a)})^n = (\cos(a) + (i\sin(a))^2)^n
And by De Moivre’s theorem we get
(cos(na)+isin(na))2(\cos(na) + i\sin(na))^2

Using the identity we derived before 1+itan(a)1−itan(a)=(cos(a)+isin(a))2\frac{1+i\tan(a)}{1-i\tan(a)} = (\cos(a) + i\sin(a))^2 we get that (1+itan(a)1−itan(a))n=1+tan(na)1−tan(na)(\frac{1+i\tan(a)}{1-i\tan(a)})^n = \frac{1 + \tan(na)}{1-\tan(na)}

  

 

Use \ before the trig name to set it properly in Latex i.e \cos (x) gives cos(x)\cos(x) which is easier to read than cos(x)cos (x)
– Mattos
2 days ago

  

 

@Mattos Sorry, didnt know you could do that, I edited my answer
– Ziad Fakhoury
2 days ago

  

 

No need to apologise, I was just giving you a tip for future reference. It looks good now. Also, if you need to put brackets around a fraction, use ‘\left (‘ and ‘\right )’ so that the brackets are big enough to fit the entire term inside.
– Mattos
2 days ago

By De Moivre,(1+itana1−itana)n=(cosa+isinacosa−isina)n=cosna+isinnacosna−isinna=1+itanna1−itanna.\left(\frac{1+i\tan a}{1-i\tan a}\right)^n=\left(\frac{\cos a+i\sin a}{\cos a-i\sin a}\right)^n=\frac{\cos na+i\sin na}{\cos na-i\sin na}=\frac{1+i\tan na}{1-i\tan na}.