## Tag Archives: roots

### Complex attitude!

Let us a discuss yet one more complex number based IITJEE mains problem.

Problem:

Let $z_{1}$ and $z_{2}$ be nth roots of unity, which subtend a right angle at the origin. Then, integer n must be of the form __________________. (fill in the blank).

Solution:

nth roots of unity are given by

$\cos{\frac{2m\pi}{n}}+ i \sin{\frac{2m\pi}{n}}= e^{\frac{2m\pi}{n}}$

where $m=0,1,2, \ldots, n-1$.

Let $z_{1}=\cos{\frac{2m_{1}\pi}{n}}+ i\sin{\frac{2m_{1}\pi}{n}}=e^{2m_{1}\pi i/n}$

Let $z_{2}=e^{2m_{2}\pi i/n}$ where $0 \leq m_{1}, m_{2}< n$, $m_{1} \neq m_{2}$

As the join of $z_{1}$ and $z_{2}$ subtends a right angle at the origin, we deduce that $\frac{z_{1}}{z_{2}}$ is purely imaginary.

$\Longrightarrow \frac{e^{2m_{1}\pi i/n}}{e^{2m_{2}\pi i/n}}=ik$, for some real k

$\Longrightarrow e^{2(m_{1}-m_{2})\pi i/n}=ik$

$\Longrightarrow n=4(m_{1}-m_{2})$. Thus, n must be of the form 4k.

More later,

Nalin Pithwa

### Intermediate value theorem

Intermediate value theorem.

Let $f:[a,b] \rightarrow \Re$ be continuous. Suppose for $x_{1},x_{2} \in [a,b]$, $f(x_{1}) \neq f(x_{2})$. If c is a real number between $f(x_{1})$ and $f(x_{2})$, then there is an $x_{0}$ between $x_{1}$ and $x_{2}$ such that $f(x_{0})=c$.

Proof.

Define $g:[a,b] \rightarrow \Re$ by $g(x)=f(x)-c$. Then, g is a continuous function such that $g(x_{1})$ and $g(x_{2})$ are of opposite signs. The assertion of the theorem amounts to saying that there is a point $x_{0}$ between $x_{1}$ and $x_{2}$ such that $g(x_{0})=0$. Without loss of generality, we may take $g(x_{1})>0$ and $g(x_{2})<0$ (otherwise replace g by -g). If $g(\frac{x_{1}+x_{2}}{2})>0$, we write $\frac{x_{1}+x_{2}}{2}=a_{1}$ and $x_{2}=b_{1}$; otherwise, write $a_{1}=x_{1}$ and $b_{1}=\frac{x_{1}+x_{2}}{2}$ so that we have $g(a_{1})>0$, $g(b_{1})<0$. Now if $g(\frac{x_{1}+x_{2}}{2})=0$, then $x_{0}=\frac{a_{1}+b_{1}}{2}$.

If $g(\frac{a_{1}+b_{1}}{2})>0$, write $\frac{a_{1}+b_{1}}{2}=a_{2}$ and $b_{1}=b_{2}$, otherwise write $a_{2}=a_{1}$ and $b_{2}=\frac{a_{1}+b_{1}}{2}$, so that we have $g(a_{2})>0$ and $g(b_{2})<0$. We could continue this process and find sequences $(a_{n})_{n=1}^{\infty}$,

$(b_{n})_{n=1}^{\infty}$ with $g(a_{n})>0$ and $g(b_{n})<0$ and

$a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq a_{n+1} \leq \ldots \leq x_{2}$,

$b_{1} \geq b_{2} \geq \ldots \geq b_{n} \geq b_{n+1} \geq \ldots \geq x_{1}$.

Since $(a_{n})_{n=1}^{\infty}$ is a monotonically non-decreasing sequence bounded above, it must converge. Suppose it converges to $\alpha$. Similarly, $(b_{n})_{n=1}^{\infty}$ is monotonically non-increasing, bounded below and therefore converges to, say, $\beta$. We further note that $b_{n}-a_{n}=\frac{x_{2}-x_{1}}{2^{n}} \rightarrow 0$ as $n \rightarrow \infty$ implying $\alpha=\beta$. Let us call this $x_{0}$. By the continuity of g, we have $\lim_{n \rightarrow \infty}g(a_{n})=g(x_{0})=\lim_{n \rightarrow \infty}g(b_{n})$, and since $g(a_{n})>0$ for all n, we must have $g(x_{0}) \geq 0$ and at the same time since $g(b_{n})<0$ for all n, we must also have $g(x_{0}) \leq 0$. This implies $g(x_{0})=0$. QED.

Corollary.

If f is a continuous function in an interval I and $f(a)f(b)<0$ for some $a,b \in I$, then there is a point c between a and b for which $f(c)=0$. (Exercise).

The above result is often used to locate the roots of equations of the form $f(x)=0$.

For example, consider the equation: $f(x) \equiv x^{3}+x-1=0$.

Note that $f(0)=-1$ whereas $f(1)=1$. This shows that the above equation has a root between 0 and 1. Now try with 0.5. $f(0.5)=-0.375$. So there must be a root of the equation between 1 and 0.5. Try $0.75$. $f(0.75)>0$, which means that the root is between 0.5 and 0.75. So, we may try $0.625$. $f(0.625)<0$. So the root is between 0.75 and 0.625. Now, if we take the approximate root to be 0.6875, then we are away from the exact root at most a distance of 0.0625. If we continue this process further, we shall get better and better approximations to the root of the equation.

Exercise.

Find the cube root of 10 using the above method correct to 4 places of decimal.

More later,

Nalin Pithwa