Question 1: Express the following complex numbers in the form
Answer:
Question 2: Find the real values of , if
Answer:
Therefore comparing, we get
… … … … … i)
… … … … … ii)
Solving i) and ii), multiplying i) by and ii) by
and adding we get
Now substituting in ii) we get
Hence and
Therefore comparing, we get
… … … … … i)
… … … … … ii)
Solving i) and ii), multiplying i) by and ii) by
and adding we get
Now substituting in i) we get
Hence and
… … … … … i)
… … … … … ii)
Solving i) and ii), multiplying i) by and ii) by
and adding we get
Substituting in i) we get
Hence and
Comparing we get
… … … … … i)
… … … … … ii)
Adding i) and ii) we get
Substituting in i) we get
Hence and
Question 3: Find the conjugates of the following complex numbers:
Answer:
Note: , then conjugate of
is
Therefore
Therefore
Therefore
Therefore
Therefore
Question 4: Find the multiplicative inverse of the following complex numbers:
Answer:
is the complex number, then multiplication inverse of
is
or
Answer:
, then
Question 6: , find
Answer:
Answer:
Answer:
Hence proved.
Therefore the general solution is given by
Question 9: Find the least positive integral value of for which
is real.
Answer:
For
For
Therefore the least positive integral value of
Answer:
For to be purely Real,
Answer:
For this to be real , the smallest positive value of will be
.
Thus , which is real.
Answer:
and
Answer:
Answer:
Answer:
Question 16:
Answer:
Now
Now,
Now
Now,
Now,
Answer:
Question 18: , then show that
Answer:
Hence proved.
Question 19: Solve the system of equations .
Answer:
Let
… … … … … i)
… … … … … ii)
Adding i) and ii) we get
Hence
For positive sign for
For negative sign for
Answer:
Let
If it is purely imaginary,
Question 21: is a complex number other than
such that
and
Answer:
Let and
… … … … … i)
Since there is no real part, is purely an imaginary numbers
Question 22: , find
Answer:
Let
Comparing,
Question 23: Solve the equation
Answer:
Let
Comparing,
and
Question 24: What is the smallest positive integer for which
Answer:
is a multiple of
.
Hence the smallest positive number for which
is
Question 25: , are complex numbers such that,
.
Answer:
Question 26: Find the number of solutions of
Answer:
Let
Now
Therefore
For
Thus there are infinite many solution of the form