## Category Archives: Complex Numbers

### Moduli and conjugates

Problem:

Let $z_{1}$, $z_{2}$, $z_{3}$ be complex numbers such that

$z_{1}+z_{2}+z_{3}=z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=0$

Prove that $|z_{1}|=|z_{2}|=|z_{3}|$.

Proof:

Method I:

Substituting $z_{1}+z_{2}=-z_{3}$ in $z_{1}z_{2}+z_{3}(z_{1}+z_{2})=0$ gives $z_{1}z_{2}=z_{3}^{2}$, so $|z_{1}|.|z_{2}|=|z_{3}|^{2}$.

Likewise, $|z_{2}|.|z_{3}|=|z_{1}|^{2}$ and $|z_{3}|.|z_{1}|=|z_{2}|^{2}$. Then,

$|z_{1}|^{2}+|z_{2}|^{2}+|z_{3}|^{2}=|z_{1}|.|z_{2}|+|z_{2}|.|z_{3}|+|z_{3}|.|z_{1}|$, that is,

$(|z_{1}|-|z_{2}|)^{2}+(|z_{2}|-|z_{3}|)^{2}+(|z_{3}|-|z_{1}|)^{2}=0$ yielding $|z_{1}|=|z_{2}|=|z_{3}|$.

Method II:

Using the relations between the roots and the coefficients, it follows that $z_{1}$, $z_{2}$, $z_{3}$ are the roots of polynomial $z^{3}-p$, where $p=z_{1}z_{2}z_{3}$. Hence, $z_{1}^{3}-p=z_{2}^{3}-p=z_{3}^{3}-p=0$ implying $z_{1}^{3}=z_{2}^{3}=z_{3}^{3}$, and the conclusion follows.

More complex fun later,

Nalin Pithwa

### Equation of a circle

Consider a fixed complex number $z_{0}$ and let $z$ be any complex number which moves in such a way that its distance from $z_{0}$ is always equal to r. This implies $z$ would lie on a circle whose centre is $z_{0}$ and radius r. And, its equation would be

$|z-z_{0}|=r$

or $|z-z_{0}|^{2}=r^{2}$

or $(z-z_{0})(\overline{z}-\overline{z_{0}})=r^{2}$,

or $z\overline{z}-z \overline{z_{0}}-\overline{z}z_{0}-r^{2}=0$

Let $-a=z_{0}$ and $z_{0}\overline{z_{0}}-r^{2}=b$. Then,

$z\overline{z}+a\overline{z}+\overline{a}z+b=0$

It represents the general equation of a circle in the complex plane.

Now, let us consider a circle described on a line segment AB $(A(z_{1}), B(z_{2}))$ as its diameter. Let $P(z)$ as its diameter. Let $P(z)$ be any point on the circle. As the angle in the semicircle is $\pi/2$, so

$\angle {APB}=\pi/2$

$\Longrightarrow (\frac{z_{1}-z}{z_{2}-z})=\pm \pi/2$

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

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

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

Condition for four points to be concyclic:

Let ABCD be a cyclic quadrilateral such that $A(z_{1})$, $B(z_{2})$, $C(z_{3})$ and $D(z_{4})$ lie on a circle. (Remember the following basic property of concyclic quadrilaterals: opposite angles are supplementary).

The above property means the following:

$\arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})+\arg (\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$

$\Longrightarrow \arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$

$(\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})$ is purely real.

Thus, points $A(z_{1})$, $B(z_{2})$, $C(z_{3})$, $D(z_{4})$ (taken in order) would be concyclic if the above condition is satisfied.

More later,

Nalin Pithwa

### 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

### A Little Note on Complex Numbers and Geometry for IITJEE Maths

Note:

1. In acute triangle, orthocentre (H), centroid (G), and circumcentre (O) are collinear and $HG:GO = 2:1$
2. Centroid of the triangle formed by points $A(z_{1})$, $B(z_{2})$, $C(z_{3})$ is $(z_{1}+z_{2}+z_{3})/3$.
3. If the circumcentre of a triangle formed by $z_{1}$, $z_{2}$ and $z_{3}$ is origin, then its orthocentre is $z_{1}+z_{2}+z_{3}$ (using 1).

Example 1:

Find the relation if $z_{1}$, $z_{2}$, $z_{3}$, $z_{4}$ are the points of the vertices of a parallelogram taken in order.

Solution:

As the diagonals of a parallelogram bisect each other, the affix of the mid-point of AC is same as the affix of the mid-point of BD. That is,

$\frac{z_{1}+z_{3}}{2}=\frac{z_{2}+z_{3}}{2}$

or $z_{1}+z_{3}=z_{2}+z_{4}$

Example 2:

if $z_{1}$, $z_{2}$, $z_{3}$ are three non-zero complex numbers such that $z_{3}=(1-\lambda)z_{1}+\lambda z_{2}$ where $\lambda \in \Re - \{ 0 \}$ then prove that the points corresponding to $z_{1}$, $z_{2}$, and $z_{3}$ are collinear.

Solution:

$z_{3}=(1-\lambda)z_{1}+z_{2} =$latex \frac{(1-\lambda)z_{1}+\lambda z_{2}}{1-\lambda +\lambda}\$.

Hence, $z_{3}$ divides the line joining $A(z_{1})$ and $B(z_{2})$ in the ratio $\lambda : (1-\lambda)$. Thus, the given points are collinear.

Homework:

1. Let $z_{1}$, $z_{2}$, $z_{3}$ be three complex numbers and a, b, c be real numbers not all zero, such that $a+b+c=0$ and $az_{1}+bz_{2}+cz_{3}=0$. Show that $z_{1}$, $z_{2}$, $z_{3}$ are collinear.
2. In triangle PQR, $P(z_{1})$, $Q(z_{2})$, and $R(z_{3})$ are inscribed in the circle $|z|=5$. If $H(z^{*})$ be the orthocentre of triangle PQR, then find $z^{*}$.

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

### Square root of a complex number

Let $a+ib$ be a complex number such that $\sqrt{a+ib} = x+iy$ where x and y are real numbers. Then,

$\sqrt{a+ib}=x+iy$

or $(a+ib)=(x+iy)^{2}$

or $a+ib=(x^{2}-y^{2})+2ixy$

On equating real and imaginary parts, we get

$x^{2}-y^{2}=a$

$2xy=b$

Now, $(x^{2}+y^{2})^{2}=(x^{2}-y^{2})^{2}+4x^{2}y^{2}$

or $(x^{2}+y^{2})^{2}=a^{2}+b^{2}$

or $(x^{2}+y^{2})=\sqrt{a^{2}+b^{2}}$ since $x^{2}+y^{2} \geq 0$

From the above, we get

$x^{2}=(1/2)(\sqrt{a^{2}+b^{2}}+a)$ and $y^{2}=(1/2)(\sqrt{a^{2}+b^{2}}-a)$

which in turn implies $x= \pm \sqrt {(1/2)(\sqrt{a^{2}+b^{2}}+a)}$

and $y=\pm \sqrt{(1/2)(\sqrt{a^{2}+b^{2}}-a)}$

If b is positive, then by the relation $2xy=b$, x and y are of the same sign. Hence,

$\sqrt{a+ib}=\pm (\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}+a)}+i\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}-a)})$

If b is negative, then by the relation $2xy=b$, x and y are of different signs. Hence,

$\sqrt{a+ib}=\pm (\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}+a)}-i\sqrt{(1/2)(\sqrt{a^{2}+b^{2}}-a)})$.

Note: When you have to actually, find the square root of a particular complex number or even a complex expression, carry out the above steps and don’t just mug up  the formula and try to substitute! There are a thousands of such derivations in math, with fancy formulae, so it is better to gain a deep understanding of the proofs rather than mug up  techniques or tips or tricks !!

Try this homework now:

1. Find the square root of $1+2i$
2. Find all possible values of $\sqrt{i}+\sqrt{-i}$
3. Solve for z: $z^{2}-(3-2i)z=(5i-5)$

Have fun the complex way 🙂

Nalin Pithwa

### 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

### Complex ain’t so complex ! Learning to think!

Problem:

If 1, $\omega, \omega^{2}, \omega^{3}, \ldots, \omega^{n}$ are the nth roots of unity, then find the value of $(2-\omega)(2-\omega^{2})(2-\omega^{3})\ldots (2-\omega^{n-1})$.

Solution:

Learning to think:

Compare it with what we know from our higher algebra — suppose we have to multiply out:

$(x+a)(x+b)(x+c)(x+d)$. We know it is equal to the following:

$x^{4}+(a+b+c+d)x^{3}+(ab+ac+ad+bc+bd+cd)x^{2}+(abc+acd+bcd+abd)x+abcd$

If we examine the way in which the partial products are formed, we see that

(1) the term $x^{4}$ is formed by taking the letter x out of each of the factors.

(2) the terms involving $x^{3}$ are formed by taking the letter x out of any three factors, in every possible way, and one of the letters a, b, c, d out of the remaining factor

(3) the terms involving $x^{2}$ are formed by taking the letter x out of any two factors, in every possible way, and two of the letters a, b, c, d out of the remaining factors

(4) the terms involving x are formed by taking the letter x out of any one factor, and three out of the letters a, b, c, d out of the remaining factors.

(5) the term independent of x is the product of all the letters a, b, c, d.

Further hint:

relate the above to sum of binomial coefficients.

and, you are almost done.

More later,

Nalin Pithwa

### Are complex numbers complex ?

You  might perhaps think that complex numbers are complex to handle. Quite contrary. They are easily applied to various kinds of engineering problems and are easily handled in pure math concepts compared to real numbers. Which brings me to another point. Mathematicians are perhaps short of rich vocabulary so they name some object as a “ring”, which is not a wedding or engagement ring at all; there is a “field”, which is not a field of maize at all; then there is a “group”, which is just an abstract object and certainly not a group of people!!

Problem:

Prove the identity:

$({n \choose 0}-{n \choose 2}+{n \choose 4}- \ldots)^{2}+({n \choose 1}-{n \choose 3}+{n \choose 5}-\ldots)^{2}=2^{n}$

Solution:

Denote

$x_{n}={n \choose 0}-{n \choose 2}+{n \choose 4}- \ldots$ and

$y_{n}={n \choose 1}-{n \choose 3}+{n \choose 5}-\ldots$

and observe that $(1+i)^{n}=x_{n}+i y_{n}$

Passing to the absolute value it follows that

$|x_{n}+y_{n}i|=|(1+i)^{n}|=|1+i|^{n}=2^{n/2}$.

This is equivalent to $x_{n}^{2}+y_{n}^{2}=2^{n}$.

More later,

Nalin Pithwa

### A bit challenging problem on complex numbers

Problem:

If $\omega$ and $\omega^{2}$ satisfy the equation

$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x} = \frac{2}{x}$

then find the value of $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1}$.

Solution:

We can write the given equation as

$\sum {x(b+x)(c+x)(d+x)}=2(a+x)(b+x)(c+x)(d+x)$

$\Longrightarrow \sum {x(x^{3}+(b+c+d)x^{2}+(bc+cd+bd)x+bcd)}$

$= 2{(x^{4}+(a+b+c+d))x^{3}+(ab+ac+ad+bc+bd+cd)x^{2}+(abc+abd+acd+bcd)x+abcd)}$

$\Longrightarrow 2x^{4}+(a+b+c+d)x^{3}+0x^{2}-(abc+abd+acd+bcd)x-2abcd=0$

This is a fourth degree equation whose two roots are $\omega, \omega^{2}$. Let $\alpha, \beta$ be the other two roots. Then,

$(\alpha + \beta)(\omega + \omega^{2}) + \alpha \beta + \omega . \omega^{2} = 0$ (sum of the products taken two at a time)

$\Longrightarrow (\alpha + \beta)(-1)+\alpha \beta +1 =0$

$\Longrightarrow (1 - \alpha)(1 - \beta) = 0$

$\alpha = 1$ or $\beta = 1$

Hence, we get $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1}=1$.

Note that in deriving the fourth degree equation we used a basic technique of expansion or multiplication of several binomials.

More later,

Nalin Pithwa