Complex Numbers « Mathematics Hothouse

Category Archives: Complex Numbers

Complex Numbers — a typical IITJEE Main problem

August 28, 2015 – 11:06 pm

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

(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!

August 26, 2015 – 2:23 pm

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

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

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

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Tagged complex numbers | Comments (0)

Simple complex problems !

August 11, 2015 – 9:38 am

Question:

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

(a) 0

(b) 2

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

(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

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam | Tagged complex numbers | Comments (0)

Minimum value

August 2, 2015 – 8:13 am

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

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam | Tagged complex numbers | Comments (0)

A Cute Complex Problem

August 1, 2015 – 6:23 am

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

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam, RMO | Tagged complex numbers, plane trigonometry problem | Comments (0)

Complex Numbers for you

July 31, 2015 – 1:59 pm

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

(a) 4

(b) 3

(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

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam, RMO | Tagged complex numbers | Comments (0)

More complex stuff

July 29, 2015 – 1:18 pm

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

(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

By Nalin Pithwa | Also posted in Cnennai Math Institute Entrance Exam, IITJEE Mains, ISI Kolkatta Entrance Exam | Tagged complex numbers | Comments (0)

De Moivre’s Theorem application

July 28, 2015 – 11:23 pm

Question:

If $f_{r}(\alpha)=(\cos{\frac{\alpha}{r^{2}}}+i\sin{\frac{\alpha}{r^{2}}}) \times (\cos{\frac{2\alpha}{r^{2}}}+i\sin{\frac{2\alpha}{r^{2}}}) \ldots (\cos{\frac{\alpha}{r}}+i\sin{\frac{\alpha}{r}})$ , then

$\lim_{n \rightarrow \infty}f_{n}{\pi}$ equals

(a) -1

(b) 1

(d) i

Solution.

Using De Moivre’s theorem,

$f_{r}{\alpha}=e^{i\frac{\alpha}{r^{2}}}e^{i\frac{2\alpha}{r^{2}}}\ldots e^{i\frac{\alpha}{r}}$

which in turn equals $e^{(i \frac{\alpha}{r^{2}})(1+2+\ldots+r)}=e^{(i\frac{\alpha}{r^{2}})(\frac{r(r+1)}{2})}=e^{i(\frac{\alpha}{2})(1+\frac{1}{r})}$

Hence, $\lim_{n \rightarrow \infty}f_{n}(\pi)=\lim_{n \rightarrow \infty}e^{(i)(\frac{\pi}{2})(1+\frac{1}{n})}=e^{i(\frac{\pi}{2})}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=i$ .