Tag Archives: complex numbers

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!!

Well, here’s your cryptic complex problem to cudgel your brains!

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

Complex Numbers — a typical IITJEE Main problem

Example 1.

If \omega is the imaginary cube root of unity, then value of the expression

1(2-\omega)(2-\omega^{2}) + 2(3-\omega)(3-\omega^{2})+\ldots+(n-1)(n-\omega)(n-\omega^{2}) is

(a) \frac{1}{4}{n^{2}}(n+1)^{2}-n

(b) \frac{1}{4}{n^{2}}(n-1)^{2}+n

(c) \frac{1}{4}{n^{2}}(n+1)-n

(d) \frac{1}{4}n(n+1)^{2}-n

Answer. (a).

Solution:

rth term of the given expression is

r(r+1-\omega)(r+1-\omega^{2})=(r+1-1)(r+1-\omega)(r+1-\omega^{2})=(r+1)^{3}-1

because x^{3}-1=(x-1)(x-\omega)(x-\omega^{2}).

Thus, the value of the expression is given by

\sum_{r=1}^{n-1}r(r+1-\omega)(r+1-\omega^{2})=\sum_{r=1}^{n-1}[(r+1)^{3}-1]

=2^{3}+3^{3}+ \ldots + n^{3}-(n-1)

=1^{3}+2^{3}+\ldots + n^{3}-n

=\frac{1}{4}n^{2}{(n+1)}^{2}-n

More later,

Nalin Pithwa

It’s complex feeling!

Example.

If z_{1}, z_{2}, z_{3} are complex numbers such that

|z_{1}|=|z_{2}|=|z_{3}|=|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=1, then |z_{1}+z_{2}+z_{3}| is

(a)  equal to 1

(b) less than 1

(c) greater than 3

(d) equal to 3

Solution:

Since |z_{1}|=|z_{2}|=|z_{3}|=1, we get z_{1} \overline{z_{1}}=z_{2} \overline{z_{2}}=z_{3} \overline{z_{3}}=1

Now, 1=|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=|\overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}|=|\overline{z_{1}+z_{2}+z_{3}}|

\Longrightarrow 1 = |z_{1}+z_{2}+z_{3}|

Simple complex problems !

Question:

If z lies on the circle |z-1|=1, then \frac{z-2}{z} equals

(a) 0

(b) 2

(c) -1

(d) none of these.

Solution:

Note that |z-1|=1 represents a circle with the segment joining z=0 and z=2+0i as a diameter. (draw this circle for yourself!)

If z lies on this circle, then \frac{z-2}{z-0} is purely imaginary.

Ans. (d).

Question:

If |z|=1 and w=\frac{z-1}{z+1} (where z \neq -1), then \Re (w) equals

(a) 0

(b) \frac{-1}{|z+1|^{2}}

(c) |\frac{z}{z+1}|\frac{1}{|z+1|^{2}}

(d) \frac{\sqrt{2}}{|z+1|^{2}}

Solution:

w=\frac{z-1}{z+1} \Longrightarrow wz+w=z-1 \Longrightarrow w+1=z(1-w) \Longrightarrow z=\frac{1+w}{1-w} \Longrightarrow |\frac{1+w}{1-w}|=|z|=1 \Longrightarrow |1+w|=|1-w|

This shows that w is equidistant from -1 and 1. Hence, w lies on the perpendicular bisector of the segment joining -1 and 1, that is, w lies on the imaginary axis.

Hence, \Re (w)=0.

More later,

Nalin Pithwa

 

 

 

Minimum value

Example. 

For any complex number z, the minimum value of |z|+|z-2i| is:

a) 0

b) 1

c) 2

d) none of these.

Solution.

We have for z \in C, |2i|=|z+(2i-z)|\leq |z|+|2i-z|

\Longrightarrow 2 \leq |z|+|z-2i|

Thus, the required minimum value is 2 and it is attained for any z lying on the segment joining z=0 and z=2i.

Answer. Option C.

More later,

Nalin Pithwa

A Cute Complex Problem

Question:

If w=\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}}, then find the value of 1+w+w^{2}+w^{3}+\ldots+w^{n-1}.

Solution:

We have S=1+w+w^{2}+w^{3}+\ldots+w^{n-1}=\frac{1-w^{n}}{1-w}.

But, w^{n}=\cos{\frac{n\pi}{n}}+i\sin{\frac{n\pi}{n}}=-1

Thus, S=\frac{2}{1-w}

but, 1-w=1-\cos{\frac{\pi}{n}}-i\sin{\frac{\pi}{n}} which equals

2\sin^{2}{\frac{\pi}{2n}}-2i\sin{\frac{\pi}{2n}}\cos{\frac{\pi}{2n}}

that is, -2i\sin{\frac{\pi}{2n}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}}].

Thus, S=\frac{-2}{2i\sin{\frac{\pi}{2n}}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}} ]^{-1}=1+i\cot{\frac{\pi}{2n}}.

Hope you are finding it useful,

More later,

Nalin Pithwa

Complex Numbers for you

If iz^{3}+z^{2}-z+i=0, then |z| equals

(a) 4

(b) 3

(c) 2

(d) 1.

Solution.

We can write the given equation as

z^{3}+\frac{1}{i}z^{2}-\frac{1}{i}z+1=0, or

z^{3}-iz^{2}+iz-i^{2}=0

\Longrightarrow z^{2}(z-i)+i(z-i)=0

\Longrightarrow (z^{2}+i)(z-i)=0 \Longrightarrow z^{2}=-i, z=i

\Longrightarrow |z|^{2}=|-i| and |z|=|i|

\Longrightarrow |z|^{2}=1 and |z|=1

\Longrightarrow |z|=1

Answer. Option d.

More later,

Nalin Pithwa

 

More complex stuff

Problem.

If z_{1}, z_{2}, \ldots , z_{n} lie on the unit circle |z|=2, then value of

E=|z_{1}+z_{2}+\ldots+z_{n}|-4|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\ldots+\frac{1}{z_{n}}| is

(a) 0

(b) n

(c) -n

(d) none of these.

Solution.

As z_{1},z_{2},\ldots, z_{n} lie on the circle |z|=2, |z_{i}|=2 \Longrightarrow |z_{i}|^{2}=4 \Longrightarrow z_{i}\overline{z_{i}}=4 for i=1,2,3, \ldots, n

Thus, \frac{1}{z_{i}}=\frac{\overline{z_{i}}}{4} for i=1, 2, 3, \ldots, n

Hence, E=|z_{1}+z_{2}+\ldots+z_{n}|-4|\frac{\overline{z_{1}}}{4}+\frac{\overline{z_{2}}}{4}+\ldots+\frac{\overline{z_{n}}}{4}|, which in turn equals

|z_{1}+z_{2}+\ldots+z_{n}|-|\overline{z_{1}}+\overline{z_{2}}+\ldots+\overline{z_{3}}|, that is,

|z_{1}+z_{2}+\ldots+z_{n}|-|\overline{z_{1}+z_{2}+\ldots+z_{n}}|=0.

(since |z|=|\overline{z}|).

Answer. Option a.

More later,

Nalin Pithwa