Complex Number
Long
Answer Question:
1.
Prove that a2 + b2 = 1 and 
=
if a
+ib
= 
Solution:
a2 +
b2
= 1 and 
=
if a +
ib = 
Let a + ib = 
=
a + ib = 
a + ib = 
a + ib = 
On Comparison,
a = , b =
Hence,
a2 +
b2 = 
a2 +
b2 =
= 1
2. Find real value of 
such that, 
is a real number
Solution:
We have 
=
= 
= 
+ 
Since = is real
= 0 = 
= 
= (2n+1) 
, n 
Z
3. Find real 
such that 
is purely real.
Solution:
z = 
z = 
z = 
z = 
z = 
+ 
For z to be purely
real
= 0
= 0
= 0, 
……
Hence = n
4. If 
= 1, show that z is real number
Solution:
Given 
= 1
= 1
= 
= 
where z =
= 
Now, 
= 
= 
= 
+25 – 10y =
+25 +10y
=0
|
|
Z=
which is real number
5. If (1+
)
(1+2)
(1+
)……..
(1+
)
= 
.
Show 2.5.10… (1+
)
= 
Solution:
We have, (1+
) (1+2) (1+
)…….. (1+
) = 
= 
=
[
= 
]
=
…(
) = 
.
6. For complex number 
=
and 
=
, find 
Solution:
= 
= 
=
= 
=
=
7. Convert complex number -3
+3
in polar form.
Solution:
Let z = -3
+3
r = 
= 
= 6
Let 
= 
= 1
= 
Since the point
representing z lies in II quadrant. Therefore, the argument of z is given by
= 
So, the polar form of z = -3 +3
is
z=
)
z=6
8. Prove that 
= (
(
(
(
Solution: RHS = (
(
(
(
=
{(
-
{(
-1
-
}
=
{(-
{(-1-
}
=
{
+2+2} {
}
=
{+2+2} {
}
=
(+2-(2
=
=
=
LHS
9. Find all the non-zero complex number
z satisfying 
= 
Solution:
Let z =
Given,
= 
= 
= 
= 0+0
…….
(i)
And 
= 0 ……
(ii)
Now 
Or 
Or 
= 
Putting
in (ii)
= 0
= 0
= 0, = 1
Thus, we have the following pairs of values of 
and
= 0,
= 0;
= 0,= 1
Putting =
in (ii) we get
= 0
= 0
0
Thus, we have the following pairs of values of 
, = and 
, = 
And z= 
Hence,
and
10. If 
+
then show that 
=1
Sol. We have, 
+
-
= 0
Or 
Now,
= =
= 1
And
= = 1
= 1
= 1
11. Find the numbers of solution of 
+
= 0
Sol. Given,
+ = 0
Let z=
=0
=0
2
= 0
2
)
= 0 or 
(not possible)
Therefore, = 0 and 
So, can have any real value. Hence, infinitely
many solution.