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