Complex Numbers Ques 109
Let $z _1$ and $z _2$ be $n$th roots of unity which subtend a right angled at the origin, then $n$ must be of the form (where, $k$ is an integer)
(2001, 1M)
(a) $4 k+1$
(b) $4 k+2$
(c) $4 k+3$
(d) $4 k$
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Answer:
Correct Answer: 109.(d)
Solution:
Formula:
Steps to determine nth roots of a complex number:
- Since, $\arg \frac{z _1}{z _2}=\frac{\pi}{2}$
$ \begin{array}{lll} \Rightarrow & \frac{z _1}{z _2} & =\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}=i \\ \therefore & \frac{z _1^{n}}{z _2^{n}} & =(i)^{n} \Rightarrow i^{n}=1 \\ \Rightarrow & n & =4 k \end{array} $
Alternate Solution
Since,
$ \arg \frac{z _2}{z _1}=\frac{\pi}{2} $
$ \therefore \quad \frac{z _2}{z _1}=\left|\frac{z _2}{z _1}\right| e^{i \frac{\pi}{2}} $
$\left[\because\left|z _1\right|=\left|z _2\right|=1\right]$
$\Rightarrow \frac{z _2}{z _1} =i $
$\Rightarrow {(\frac{z _2}{z _1})}^{n} =i^{n}$
$\therefore \quad z _1$ and $z _2$ are $n$th roots of unity.
$ z _1^{n}=z _2^{n}=1 $
$\Rightarrow \quad (\frac{z _2}{z _1})^n=1$
$\Rightarrow \quad i^{n}=1$
$\Rightarrow \quad n=4 k$, where $k$ is an integer.
BETA
