Category Archives: Complex Numbers

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

 

De Moivre’s Theorem application

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

(c) -i

(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.

More complex stuff to be continued in next blog (pun intended) 🙂

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.