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Given $z\in\mathbb{C}$ with $|z|=1$, then prove that the equation $\Big(\dfrac{1+ia}{1-ia}\Big)^4=z$ has all roots real and distinct

My Attempt $$ z=e^{i\theta}\implies z^{1/4}=e^{i\theta/4}=\frac{1+ia}{1-ia}.1^{1/4}=\frac{1+ia}{1-ia}.(1,-1,i,-i) $$ Is it correct to introduce the quadratic root of unity for the complete solution ?

Case 1: $1^{1/4}=1$ $$ \frac{1}{ia}=\frac{e^{i\theta/4}+1}{e^{i\theta/4}-1}=\frac{2\cos^2\theta/8+2i\sin\theta/8.\cos\theta/8}{-2\sin^2\theta/8+2i\sin\theta/8.\cos\theta/8}\\ \frac{1}{ia}=i\cot\theta/8\\\boxed{a=-\tan\theta/8} $$ Case 2: $1^{1/4}=-1$ $$ \frac{1}{ia}=\frac{-e^{i\theta/4}+1}{-e^{i\theta/4}-1}=\frac{2\sin^2\theta/8-2i\sin\theta/8.\cos\theta/8}{-2\cos^2\theta/8-2i\sin\theta/8.\cos\theta/8}\\ \frac{1}{ia}=i\tan\theta/8\implies\boxed{a=-\cot\theta/8} $$ Case 3: $1^{1/4}=-i$ $$ \frac{1}{ia}=\frac{ie^{i\theta/4}+1}{ie^{i\theta/4}-1}=\frac{1-\sin\theta/4+i\cos\theta/4}{-1-\sin\theta/4+i\cos\theta/4}\\ =\frac{1-\cos(\pi/2-\theta/4)+i\sin(\pi/2-\theta/4)}{-1-\cos(\pi/2-\theta/4)+i\sin(\pi/2-\theta/4)}\\ =\frac{2\sin^2(\pi/4-\theta/8)+2i\sin(\pi/4-\theta/8)\cos(\pi/4-\theta/8)}{-2\cos^2(\pi/4-\theta/8)+2i\sin(\pi/4-\theta/8)\cos(\pi/4-\theta/8)}\\ \frac{1}{ia}=-i\tan(\pi/4-\theta/8)\\ \boxed{a=\cot(\pi/4-\theta/8)=\tan(\pi/4+\theta/8)} $$ Case 4: $1^{1/4}=i$ $$ \frac{1}{ia}=\frac{-ie^{i\theta/4}+1}{-ie^{i\theta/4}-1}=\frac{1+\sin\theta/4-i\cos\theta/4}{-1+\sin\theta/4-i\cos\theta/4}\\ =\frac{1+\cos(\pi/2-\theta/4)-i\sin(\pi/2-\theta/4)}{-1+\cos(\pi/2-\theta/4)-i\sin(\pi/2-\theta/4)}\\ =\frac{2\cos^2(\pi/4-\theta/8)-2i\sin(\pi/4-\theta/8)\cos(\pi/4-\theta/8)}{-2\sin^2(\pi/4-\theta/8)-2i\sin(\pi/4-\theta/8)\cos(\pi/4-\theta/8)}\\ =i.\cot(\pi/4-\theta/8)\\ \frac{1}{ia}=i\tan(\pi/4+\theta/8)\\ \boxed{a=-\cot(\pi/4+\theta/8)} $$ Is it the right way to approach the problem ? and whats the easiest way to solve this ?

Sooraj S
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2 Answers2

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For the reality of the roots, take module on both sides : $$\left|\frac{i-a}{-i-a}\right|=|z|^{1/4}=1$$ so if $A$ is the point of affix $a$, $I$ of affix $i$ and $I'$ of affix $-i$, this says that $AI=AI'$, so $A$ belongs to the perpendicular bisector of segment $[II']$, therefore the real axis.

For distinct roots, take argument : $(\overrightarrow{AI'};\overrightarrow{AI})=\frac14\arg(z)+k\frac\pi2$ with $k\in\{0,1,2,3\}$, this means 4 distinct points.

(There could be a better argument for last point).

With some calculus (which doesn't explain why roots are real) : let $z=e^{i\theta}$, and for $k\in\{0,1,2,3\}$, $\theta_k=\theta/4+k\pi/2$. Equation solves as $$a=\frac{e^{i\theta_k}-1}{i(e^{i\theta_k}+1)} = \frac{e^{i\theta_k/2}-e^{-i\theta_k/2}}{i(e^{i\theta_k/2}+e^{-i\theta_k/2})} = \tan(\theta_k/2) = \tan(\theta/8+k\pi/4)$$ OK, so it's real, but why ?

Nicolas FRANCOIS
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  • So u mean $$ \frac{1+ia}{1-ia}=z^{1/4}1^{1/4}=e^{i\theta/4}(1,-1,i,-i)\ ia=\frac{e^{i\theta/4}(1,-1,i,-i)-1}{e^{i\theta/4}(1,-1,i,-i)+1}=\frac{e^{i\theta_k}-1}{e^{i\theta_k}+1}\\theta_k=\theta/4+\arg(1,-1,i,-i)=\theta/4+(0,\pi,\pi/2,3\pi/2)\ ia=\frac{e^{i\theta_k/2}-e^{-i\theta_k/2}}{e^{i\theta_k/2}+e^{-i\theta_k/2}}=\frac{2i\sin(\theta_k/2)}{2\cos(\theta_k/2)}=i\tan(\theta_k/2)\ a=\tan(\theta_k/2)=\begin{cases} \tan(\theta/8)\ \tan(\theta/8+\pi/2)=-\cot(\theta/8)\ \tan(\theta/8+\pi/4)\ \tan(\theta/8+3\pi/4)=-\cot(\theta/8+\pi/4) \end{cases} $$ right ? – Sooraj S Mar 05 '19 at 17:58
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    Sort of. I don't like your "notation" $(1,i,-1,-i)$, and would certainly not use it with my students, but apart from this, I'm OK with what you do. – Nicolas FRANCOIS Mar 05 '19 at 19:23
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Observe that $$\left|\dfrac{1+ia}{1-ia}\right|=1$$

WLOG $$\dfrac{1+ia}{1-ia}=\cos2t+i\sin2t=\dfrac{\cos t+i\sin t}{\cos t-i\sin t}$$ where $t$ is real

Using https://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm,

$a=\tan t$

$\implies z=\cos8t+i\sin8t$

Now using repeated root theorem like How to solve $x^3=-1$?,

$y^4=z$ can not have repeated roots

lab bhattacharjee
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