## Tag Archives: algebra for IITJEE

### Equation of a line : Geometry and Complex Numbers

Equation of the line passing through the point $z_{1}$ and $z_{2}$:

Ref: Mathematics for Joint Entrance Examination JEE (Advanced), Second Edition, Algebra, G Tewani.

There are two forms of this equation, as given below:

$\left | \begin{array}{ccc} z & \overline{z_{1}} & 1 \\ z_{1} & \overline{z_{2}} & 1 \\ z_{2} & \overline{z_{2}} & 1 \end{array} \right |=0$

and $\frac{z-z_{1}}{\overline{z}-\overline{z_{1}}}=\frac{z_{1}-z_{2}}{\overline{z_{}}-\overline{z_{2}}}$

Proof:

Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$. Let A and B be the points representing $z_{1}$ and $z_{2}$ respectively.

Let $P(z)$ be any point on the line joining A and B. Let $z=x+iy$. Then $P \equiv (x,y)$, $A \equiv (x_{1}, y_{1})$ and $B \equiv (x_{2},y_{2})$. Points P, A, and B are collinear.

See attached JPEG figure 1.

The figure shows that the three points A, P  and B are collinear.

Shifting the line AB at the origin as shown in the figure; points O, P, Q are collinear. Hence,

$\arg(z-z_{2})=\arg(z_{1}-z_{2})$ or

$\arg {\frac{z-z_{2}}{z_{1}-z_{2}}}=0$

$\Longrightarrow \frac{z-z_{2}}{z_{1}-z_{2}}$ is purely real.

$\frac{z-z_{2}}{z_{1}-z_{2}}=\frac{\overline{z-z_{2}}}{z_{1}-z_{2}}$

or, $\frac{z-z_{2}}{z_{1}-z_{2}}=\frac{\overline{z}-\overline{z_{2}}}{\overline{z_{1}}-\overline{z_{2}}}$ call this as Equation 1.

$\left | \begin{array}{ccc} z & \overline{z} & 1 \\ z_{1} & \overline{z_{1}} & 1\\ z_{2} & \overline{z_{2}} & 1 \end{array} \right |=0$. Call this as Equation 2.

Hence, from (2), if points $z_{1}$, $z_{2}$, $z_{3}$ are collinear, then

$\left | \begin{array}{ccc} z_{1} & \overline{z_{1}} & 1 \\ z_{2} & \overline{z_{2}} & 1 \\ z_{3} & \overline{z_{3}} & 1 \end{array} \right |=0$.

Equation (2) can also be written as

$(\overline{z_{1}}- \overline{z_{2}}) - (z_{1}-z_{2})\overline{z}+z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}=0$

$\Longrightarrow i(\overline{z_{1}}-\overline{z_{2}})z-(z_{1}-z_{2})\overline{z} + z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}=0$

$\Longrightarrow \overline{a}z + a\overline{z}+b=0$ let us call this Equation 3.

where $a=-i(z_{1}-z_{2})$ and $b=i(z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}})=i 2i \times \Im (z_{1} \overline{z_{2}})$, which in turn equals

$-2 \times \Im(z_{1}\overline{z_{2}})$, which is a real number.

Slope  of the given line

In Equation (3), replacing z by $x+iy$, we get $(x+iy)\overline{a} + (x-iy)a+b=0$,

$\Longrightarrow (a+\overline{a})x + iy(\overline{a}-a)+b=0$

Hence, the slope $= \frac{a+\overline{a}}{i(a-\overline{a})}=\frac{2 \Re(a)}{2i \times \Im(a)}=-\frac{\Re(a)}{\Im(a)}$

Equation of a line parallel to the line $z \overline{a}+\overline{z}a+b=0$ is $z \overline{a} + \overline{z} a + \lambda=0$ (where $\lambda$ is a real number).

Equation of a line perpendicular to the line $z\overline{a}+\overline{z}a+b=0$ is $z\overline{a}+\overline{z} a + i \lambda=0$ (where $\lambda$ is a real number).

Equation of a perpendicular bisector

Consider a line segment joining $A(z_{1})$ and $B(z_{2})$. Let the line L be its perpendicular bisector. If $P(z)$ be any point on L, then we have (see attached fig 2)

$PA=PB \Longrightarrow |z-z_{1}|=|z-z_{2}|$

or $|z-z_{1}|^{2}=|z-z_{2}|^{2}$

or $(z-z_{1})(\overline{z}-\overline{z_{1}})=(z-z_{2})(\overline{z}-\overline{z_{2}})$

or

Here, $a= z_{2}-z_{1}$ and $b=z_{1}\overline{z_{1}}-z_{2} \overline{z_{2}}$

Distance of a given point from a given line:

(See attached Fig 3).

Let the given line be $z \overline{a} + \overline{z} a + b=0$ and the given point be $z_{c}$. Then,

$z_{c}=x_{c}+iy_{c}$

Replacing z by $x+iy$ in the given equation, we get

$x(a+\overline{a})+iy(\overline{a}-a)+b=0$

Distance of $(x_{c},y_{c})$ from this line is

$\frac{|x_{c}(a+\overline{a})+iy_{c}(\overline{a}-a)+b|}{\sqrt{(a+\overline{a})^{2}-(a-\overline{a})^{2}}}$

which in turn equals

$\frac{z_{c}\overline{a}+\overline{z_{c}}a+b}{\sqrt{4(\Re(a))^{2}+4(\Im(a))^{2}}}$ which is equal to finally

$\frac{|z_{c}\overline{a}+\overline{z_{c}}a+b|}{2|a|}$.

More later,

Nalin Pithwa

### Geometry with complex numbers — section formula

It ain’t complex, it’s simple !!

Section Formula:

If $P(z)$ divides the line segment joining $A(z_{1})$ and $B(z_{2})$ internally in the ratio $m:n$, then

$z = \frac{mz_{2}+nz_{1}}{m+n}$

If the division is external, then $z=\frac{mz_{2}-nz_{1}}{m-n}$

Proof:

Let $z_{1}=x_{1}+iy_{1}$, $z_{2}=x_{2}+iy_{2}$. Then, $A \equiv (x_{1},y_{1})$ and $B \equiv (x_{2},y_{2})$.

Let $z = x+iy$. Then, $P \equiv (x,y)$. We know from co-ordinate geometry,

$x = \frac{mx_{2}+nx_{1}}{m+n}$ and $y=\frac{my_{2}+my_{1}}{m+n}$

Hence, complex number of P is

$z = \frac{mx_{2}+nx_{1}}{m+n}+i\frac{my_{2}+my_{1}}{m+n}$

$\frac{m(x_{2}+iy_{2})+n(x_{1}+iy_{1})}{m+n}$

$mz_{2}+nz_{1}$

more later,

Nalin Pithwa

### A complex equation

Find the number of solutions of the equation $z^{3}+\overline{z}=0$.

Solution.

Given that $z^{3}+\overline{z}=0$. Hence, $z^{3}=-z$.

$|z|^{3} =|-\overline{z}| \Longrightarrow |z|^{3}=|z|$.Hence, we get

$|z|(|z|-1)(|z|+1)=0 \Longrightarrow |z|=0, |z|=1$ (since $|z|+1>0$)

If $|z|=1$, we get $|z|^{2}=1 \Longrightarrow z.\overline{z}=1$.

Thus, $z^{3}+\overline{z}=0 \Longrightarrow z^{3}+1/z=0$

Thus, $z^{4}+1=0 \Longrightarrow z^{4}=\cos{\pi}+i\sin{\pi}$, that is,

$z=\cos{\frac{2k+1}{4}}\pi+i\sin{\frac{2k+1}{4}}\pi$ for $k=0,1,2,3$. Therefore, the given equation has five solutions.

### Basic Algebra for IITJEE Main and RMO

More basic algebra for you guys who are thirsting for more…The following is a nice problem indicating some basic concepts or tricks in problems involving logarithms/powers.

Solve for x: $4^{x}-3^{x-(1/2)}=3^{x+(1/2)}-2^{2x-1}$ (IITJEE 1978)

Solution:

Writing $4^{x}$ as $2^{2x}$ and bringing powers of the same number on the same side we get,

$2^{2x}+2^{2x-1}=3^{x+(1/2)}+3^{x-(1/2)}$

The first term on the LHS can be written as $2^{2x-1} \times 2$, and hence, a common factor of

$2^{2x-1}$ comes out from the terms on the LHS. As for RHS, we can rewrite the first term as

$3^{x-(1/2)} \times 3$ and then the factor $3^{x-(1/2)}$ comes out as common. So, we get

$2^{2x-1}(2+1)=3^{x-(1/2)}(3+1)$, that is, $3 \times 2^{2x-1}=4 \times 3^{x-(1/2)}$

Bringing all powers of 2 to  the left and all powers of 3 to the right, we get

$2^{2x-3}=3^{x-(3/2)}$

By inspection, $x=3/2$ is a solution. But, how do we arrive at it systematically? Also, how do we know that there is no other solution? It is tempting to try to do this by saying that a power of 2 can equal a power of 3 only when are both are equal to 1. (Such a reasoning is indeed useful in solving equations in Number Theory where we mostly deal with positive integers and their factorization into integers). But, here it is inapplicable because we do not  know that the exponents are integers. Instead, let us express both the sides as the power of the same number. One way to do this  is to write 3 as $2^{\log_{2}3}$ in the RHS. Then, we can get

$2^{2x-3}=2^{(\log_{2}3)(x-(3/2)}$

As the bases are the same, the equality of powers implies that of the exponents. So, we have

$2x-3=(\log_{2}3)(x-(3/2))$

This can be solved easily to give $x=\frac{3-(3/2)(\log_{2}3)}{2-\log_{2}3}$

which simply equals $3/2$.

Hence, $x=3/2$ is the only solution.

### A variant of mathematical induction — for IITJEE Math and RMO Math

As is generally known, the principal of mathematical induction is used in the following form: We are supposed to prove a proposition for natural numbers. We check whether it is true for the natural number 1, and then assume it to be true for another natural number m. Then, we got to  show  that the proposition also holds for the natural number $m+1$.

Here is a variation of the principle of mathematical induction:

Assume the proposition to be true for a first natural number k (you have to  find the particular value of k under the conditions of the given problem). Then, assume it to be true for the natural number m. Finally, we gotta show that the proposition is true for the natural number $m+1$.

Example. Show that $n! > 3^{n}$ if n is large enough.

Solution. How large is large? We have to find that “first” value for which the above proposition is true. We actually need to check whether the inequality holds for $n=1, 2, 3, 4, 5, 6, 7, 8 \ldots$

So, if we tabulate the values of LHS and RHS, we find that the first natural number for which this holds is $n=7$..

Hence, the proposition $n! > 3^{n}$ is true for n=7.

Assume the proposition to be true for some natural number m where $m > 7$.

That is, $m! > 3^{m}$ holds true. Multiplying both sides by $m+1$, we get

$(m+1)m! > 3^{m} (m+1)$ that is,

$(m+1)! > m \times 3^{m} + 3^{m}$, but $m>7$. Hence, we get

$(m+1)! > 7 \times 3^{m} + 3^{m}$ which equals $8 \times 3^{m}$.

Hence, obviously, $(m+1)! > 3^{m} \times 3$ that is,

$(m+1)! > 3^{m+1}$. That is, the proposition holds for all natural numbers. Thus, the proof.

More later,

Nalin Pithwa

### Reciprocal equation for IITJEE and RMO/INMO

In this set of little exercises, you will get a grip on reciprocal equations.

reciprocal polynomial has the form

$ax^{n}+bx^{n-1}+cx^{n-2}+...+cx^{2}+bx+a$

in which $a \neq 0$ and the coefficients are symmetric about the middle one. A reciprocal equation is of the form $p(t)=0$ with $p(t)$ a reciprocal polynomial.

1(a) Verify that each of the following polynomials is a reciprocal polynomial:

$x^{3}+4x^{2}+4x+1$

$3x^{6}-7x^{5}+5x^{4}+2x^{3}+5x^{2}-7x+3$

1(b) Show that 0 is not a zero of any reciprocal polynomial.

1(c) Show that -1 is a zero of any reciprocal polynomial of odd degree, and deduce that any reciprocal polynomial of odd degree can be written in the form $(x+1)q(x)$, with $q(x)$ a reciprocal polynomial of even degree.

1(d) Show that, if r is a root of a reciprocal equation, then so also is $1/r$.

2(a) Let $ax^{2k}+bx^{2k-1}+...+rx^{k}+...+bx+a$ be a reciprocal equation of even degree $2k$. Show that this equation can be rewritten as

$a(x^{k}+x^{-k})+b(x^{k-1}+x^{-k+1})+...+r=0$

2(b) Let $t=x+x^{-1}$. Verify that $x^{2}+x^{-2}=t^{2}-2$ and that $x^{3}+x^{-3}=t^{3}-3t$. Prove that, in general, $x^{m}+x^{-m}$ is a polynomial of degree m in t.

2(c) Use the substitution in 2b to show that the reciprocal equation in 2a can be rewritten as an equation of degree k in the variable t. Deduce that the solution of a reciprocal equation of degree $2k$ can in general be reduced to solving one polynomial equation of degree k as well as at most k quadratic equations.

3(a) Show that a product of reciprocal polynomials is a reciprocal polynomial.

3(b) Show that, if f, g,  h are polynomials with $f=gh$ and f and h are both reciprocal polynomials, then g is also a reciprocal polynomial.