August 28, 2015 – 11:06 pm
Example 1.
If
is the imaginary cube root of unity, then value of the expression
is
(a) 
(b) 
(c) 
(d) 
Answer. (a).
Solution:
rth term of the given expression is

because
.
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]](https://s0.wp.com/latex.php?latex=%5Csum_%7Br%3D1%7D%5E%7Bn-1%7Dr%28r%2B1-%5Comega%29%28r%2B1-%5Comega%5E%7B2%7D%29%3D%5Csum_%7Br%3D1%7D%5E%7Bn-1%7D%5B%28r%2B1%29%5E%7B3%7D-1%5D&bg=ffffff&fg=000&s=0&c=20201002)



More later,
Nalin Pithwa
August 26, 2015 – 2:23 pm
Example.
If
are complex numbers such that
, then
is
(a) equal to 1
(b) less than 1
(c) greater than 3
(d) equal to 3
Solution:
Since
, we get 
Now, 

August 11, 2015 – 9:38 am
Question:
If z lies on the circle
, then
equals
(a) 0
(b) 2
(c) -1
(d) none of these.
Solution:
Note that
represents a circle with the segment joining
and
as a diameter. (draw this circle for yourself!)
If z lies on this circle, then
is purely imaginary.
Ans. (d).
Question:
If
and
(where z \neq -1), then
equals
(a) 0
(b) 
(c) 
(d) 
Solution:

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,
.
More later,
Nalin Pithwa
Example.
For any complex number z, the minimum value of
is:
a) 0
b) 1
c) 2
d) none of these.
Solution.
We have for
, 

Thus, the required minimum value is 2 and it is attained for any z lying on the segment joining
and
.
Answer. Option C.
More later,
Nalin Pithwa
Question:
If
, then find the value of
.
Solution:
We have
.
But, 
Thus, 
but,
which equals

that is,
.
Thus,
.
Hope you are finding it useful,
More later,
Nalin Pithwa
If
, then
equals
(a) 4
(b) 3
(c) 2
(d) 1.
Solution.
We can write the given equation as
, or



and 
and 

Answer. Option d.
More later,
Nalin Pithwa
Problem.
If
lie on the unit circle
, then value of
is
(a) 0
(b) n
(c) -n
(d) none of these.
Solution.
As
lie on the circle
,
for 
Thus,
for 
Hence,
, which in turn equals
, that is,
.
(since
).
Answer. Option a.
More later,
Nalin Pithwa
Question:
If
, then
equals
(a) -1
(b) 1
(c) -i
(d) i
Solution.
Using De Moivre’s theorem,

which in turn equals 
Hence,
.
More complex stuff to be continued in next blog (pun intended) 🙂
Nalin Pithwa
Find the number of solutions of the equation
.
Solution.
Given that
. Hence,
.
.Hence, we get
(since
)
If
, we get
.
Thus, 
Thus,
, that is,
for
. Therefore, the given equation has five solutions.