Referring to the AP Inter 1st Year Maths Study Material Chapter 4 Complex Numbers and Quadratic Equations Exercise 4c Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Complex Numbers and Quadratic Equations Solutions Exercise 4c
I. Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.
Question 1.
z = – 1 – i√3
Solution:
Given that z = -1 – i√3
Let r cos θ = -1 and r sin θ = -√3
Squaring and adding, we obtain (rcos θ)2 + (rsin θ)2 = (-1)2 + (-√3)2
⇒ r2 (cos θ)2 + (rsin θ)2 = (-1)2 + (√3)2
⇒ r2(cos2θ + sin2θ) = 1 + 2
⇒ r2 = 4 [∵ cos2θ + sin2θ = 1]
⇒ r = √4 = 2 [∵ Conventionally, r > 0]
Therefore, Modulus = 2
Hence, 2cos θ = -1 and 2 sin θ = -√3
⇒ cos θ = \(\frac{1}{2}\) and sin θ = –\(\frac{\sqrt{3}}{2}\)
Since both the values of sin 0 and cos 0 are negative in III quadrant.
Argument = -(π – \(\frac{\pi}{3}\)) = \(\frac{-2 \pi}{3}\)
Thus, the modulus and argument of the complex number -1 – i√3 are 2 and \(\frac{-2 \pi}{3}\) respectively.
Question 2.
z = -√3 + i
Solution:
Given that z = -√3 + i
Let r cos θ = -√3 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2θ + r2 sin2θ = (-√3)2 + 12
⇒ r2 = 3 + 1 = 4 [∵ cos2θ + sin2θ = 1]
⇒ r = √4 = 2 [∵ Conventionally, r > 0]
Therefore, Modulus = 2, Hence, 2 cos θ = -√3 and 2 sin θ = 1
⇒ cos θ = \(-\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)
Since, θ lies in the quadrant II, θ = π – \(\frac{\pi}{6}=\frac{5 \pi}{6}\)
Thus, the modulus and argument of the complex number – √3 + i are 2 and \(\frac{5 \pi}{3}\) respectively.
Convert each of the complex numbers given in exercises 3 to 8 in the polar form.
Question 3.
1 – i
Solution:
Given that z = 1 – i.
Let r cos θ = 1 and r sin θ = -1
On squaring and adding, we obtain
r2 cos2θ + r2 sin2θ = 12 + (-1)2
⇒ r2 (∵ cos2θ + sin2θ) = 1 + 1
⇒ r2 = 2
⇒ r = √2 [∵ Conventionally, r > 0]
Since, 0 lies in the IV quadrant θ = \(\left(\frac{-\pi}{4}\right)\)
Hence, 1 – i = r cos θ + ir sin θ
= √2cos \(\left(\frac{-\pi}{4}\right)\) + i sin\(\left(\frac{-\pi}{4}\right)\)
= √2[cos\(\left(\frac{-\pi}{4}\right)\) + isin\(\left(\frac{-\pi}{4}\right)\)]
Thus, this is the required polar form.
Question 4.
-1 + i
Solution:
Given that z = -1 + i.
Let r cos θ = -1 and r sin θ = 1
On squaring and adding, we obtain r2 cos2θ + r2 sin2θ = (- 1)2 + 12
⇒ r2(cos2θ + sin2θ) = 1 + 1
⇒ r2 = 2
⇒ r = √2 [∵ Conventionally, r > 0]
Therefore,√2cos θ = -1 and √2 sin θ = 1
⇒ cos θ = –\(\frac{1}{\sqrt{2}}\) and sin θ = \(\frac{1}{\sqrt{2}}\)
Since, θ lies in the II quadrant, θ = π – \(\frac{\pi}{4}=\frac{3 \pi}{4}\)
⇒ r2 (cos2θ + sin2θ) = 1 + 1
Hence, -1 – i = r cos θ + i r sin θ = √2 cos \(\frac{-3 \pi}{4}\) + i √2 sin \(\frac{-3 \pi}{4}\)
= √2(cos \(\frac{-3 \pi}{4}\) + i sin \(\frac{-3 \pi}{4}\))
Thus, this is the required polar form.
Question 5.
-1 -i ,
Solution:
Given that z = -1 – i.
Let r cos θ = – 1 and r sin θ = -1
On squaring and adding, we obtain
r2 cos2 θ + r2 sin2θ = (- 1)2 + (- 1)2
Therefore, √2 cos θ = -1 and √2 sin θ = -1
⇒ cos θ = \(\\frac{-1}{\sqrt{2}}\) and sin θ = \(\frac{-1}{\sqrt{2}}\)
Hence, -1 – i = r cos θ + i r sin θ
= √2 cos \(\frac{-3 \pi}{4}\) + i 2 sin\(\frac{-3 \pi}{4}\)
= √2(cos \(\frac{-3 \pi}{4}\) + is sin \(\frac{-3 \pi}{4}\))
Thus, this is the required polar form.
Question 6.
-3
Solution:
Given that z = -3.
Let r cos θ = -3 and r sin θ = 0
On squaring and adding, we obtain
r2 cos2θ + r2 sin2θ = (- 3)2 + (0)2
⇒ r2 (cos2θ + sin2θ) = 9 + 0
⇒ r2 = 9
⇒ r = 3 [∵ Conventionally, r > 0]
Therefore, 3 cos θ = – 3 and 3 sin θ = 0
⇒ cos θ = – 1 and sin θ = 0.
Since, θ lies in the II quadrant, θ = n
Hence, – 3 = r cos θ + ir sin θ = 3 cos π + i3 sin π = 3(cos π + i sin π)
Thus, this is the required polar form.
Question 7.
√3 + i
Solution:
Given that z = √3 + i
Let r cos θ = √3 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2θ + r2 sin2θ = (^3 )2 + 12
⇒ r2 (cos20 + sin20) = 3+l
⇒ r2 = 4
⇒ r = √4 = 2 [∵ Conventionally, r > 0]
Therefore, 2 cos θ = √3 and 2 sin θ = 1
⇒ cos θ = \(\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)
Since, θ lies in I quadrant, θ = \(\frac{\pi}{6}\)
Hence, √3+ 1 = r cos θ + ir sin θ = 2cos \(\frac{\pi}{6}\) + i2 sin \(\frac{\pi}{6}\) = 2(cos \(\frac{\pi}{6}\) + isin \(\frac{\pi}{6}\))
Thus, this is the required polar form.
Question 8.
i
Solution:
Given that z = i
Let r cos θ = 0 and r sin θ = 1
On squaring and adding, we obtain
r2 cos2θ + r2 sin2θ = (√3 )2 + 12
⇒ r2 (cos2θ + sin2θ) = 1
⇒ r2 = 1
⇒ r = √1 =1 [∵ Conventionally, r > 0]
Therefore, cos θ = 0 and sin θ = 1
Since, θ lies in the I quadrant, θ = \(\frac{\pi}{2}\)
Hence, i = r cos θ + ir sin θ = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
Thus, this is the required polar form.