Exercise 5.2 (Complex number)
Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.(Exercise 5.2 complex no. ncert math solution class 11)
Question 1:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
are negative but
is positive. So,
lies in III quadrant. Therefore,
Argument
Question 2:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
are negative but
is positive. So,
lies in II quadrant. Therefore,
Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:
Question 3:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
are negative but
is positive. So,
lies in IV quadrant. Therefore,
Therefore, the polar form of is
Question 4:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
are negative but
is positive. So,
lies in II quadrant. Therefore,
Therefore, the polar form of is
Question 5:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
are negative but tan
is positive. So,
lies in III quadrant. Therefore,
Therefore, the polar form of is given by
Question 6:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
are 0 but
is negative. Therefore,
Therefore, the polar form of is
Question 7:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (2) by (1), we have
From the equations (1), (2) and (3), it is clear that and
all are positive. So,
lies in I quadrant. Therefore,
Therefore, the polar form of is
Question 8:
Solution : Let
Comparing the real and imaginary parts, we have
Squaring and adding equation (1) and (2), we have
Therefore, modulus
Now, dividing equation (1) by (2), we have
From the equations (1), (2) and (3), it is clear that and
are 0 but
is positive. Therefore,
Therefore, the polar form of is given by