OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Exercise 9 (D)

Find the modulus and amplitude of the following complex numbers and hence express them into polar form.

 

Question 1. \( \sqrt{3} + i \)
Answer: Let the complex number be \( z = \sqrt{3} + i \). We can write this as \( x + iy \), where \( x = \sqrt{3} \) and \( y = 1 \). Since both \( x \) and \( y \) are positive, the point \( (x, y) = (\sqrt{3}, 1) \) lies in the first quadrant. This means the argument will be equal to the principal angle.
First, calculate the modulus (distance from origin):
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{3})^2 + (1)^2} \)
\( \implies r = \sqrt{3 + 1} = \sqrt{4} = 2 \)
Next, calculate the principal argument:
Let \( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{1}{\sqrt{3}} \right| \)
\( \implies \alpha = \frac{\pi}{6} \)
Since \( z \) is in the first quadrant, its argument is \( \arg(z) = \alpha = \frac{\pi}{6} \).
Therefore, the polar form of \( z \) is given by \( z = r(\cos \alpha + i \sin \alpha) \).
\( \implies z = 2\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right) \)
In simple words: We first find how far the number is from zero (modulus) and then the angle it makes with the positive x-axis (argument). Since both parts of the number are positive, the angle is straightforward.

🎯 Exam Tip: Always identify the quadrant of the complex number first. This helps determine the correct formula for the argument (e.g., \( \pi - \alpha \), \( -\alpha \), or \( -(\pi - \alpha) \)).

 

Question 2. \( - \sqrt{3} + i \)
Answer: Let the complex number be \( z = - \sqrt{3} + i \). We write this as \( x + iy \), where \( x = - \sqrt{3} \) and \( y = 1 \). Since \( x \) is negative and \( y \) is positive, the point \( (x, y) = (- \sqrt{3}, 1) \) lies in the second quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(- \sqrt{3})^2 + (1)^2} \)
\( \implies r = \sqrt{3 + 1} = \sqrt{4} = 2 \)
Next, calculate the principal angle \( \alpha \):
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{1}{- \sqrt{3}} \right| = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
\( \implies \alpha = \frac{\pi}{6} \)
Since \( z \) is in the second quadrant, its argument is \( \arg(z) = \pi - \alpha \).
\( \implies \arg(z) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)
Therefore, the polar form of \( z \) is given by \( z = r(\cos \theta + i \sin \theta) \).
\( \implies z = 2\left(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}\right) \)
In simple words: This number has a negative real part and a positive imaginary part, placing it in the top-left section of the complex plane. We find its distance from the origin and then calculate its angle, which is \( \pi \) (180 degrees) minus the reference angle.

🎯 Exam Tip: When the complex number is in the second or third quadrant, remember to adjust the reference angle \( \alpha \) by adding or subtracting it from \( \pi \) to get the correct argument.

 

Question 3. \( - 2 + 2\sqrt{3} i \)
Answer: Let the complex number be \( z = - 2 + 2\sqrt{3} i \). We set \( x = - 2 \) and \( y = 2\sqrt{3} \). Since \( x \) is negative and \( y \) is positive, the point \( (x, y) = (- 2, 2\sqrt{3}) \) lies in the second quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (2\sqrt{3})^2} \)
\( \implies r = \sqrt{4 + 12} = \sqrt{16} = 4 \)
Next, calculate the principal angle \( \alpha \):
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{2\sqrt{3}}{-2} \right| = \tan^{-1} (\sqrt{3}) \)
\( \implies \alpha = \frac{\pi}{3} \)
Since \( z \) is in the second quadrant, its argument is \( \arg(z) = \pi - \alpha \).
\( \implies \arg(z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \)
Therefore, the polar form of \( z \) is given by \( z = r(\cos \theta + i \sin \theta) \).
\( \implies z = 4\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) \)
In simple words: This complex number is in the second quadrant because its real part is negative and its imaginary part is positive. We find its length from the origin and its angle by subtracting the reference angle from \( \pi \).

🎯 Exam Tip: Always be careful with the signs of the real and imaginary parts to correctly identify the quadrant, which is essential for calculating the argument.

 

Question 4. \( 1 - i \)
Answer: Let the complex number be \( z = 1 - i \). We have \( x = 1 \) and \( y = -1 \). Since \( x \) is positive and \( y \) is negative, the point \( (x, y) = (1, -1) \) lies in the fourth quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (-1)^2} \)
\( \implies r = \sqrt{1 + 1} = \sqrt{2} \)
Next, calculate the principal angle \( \alpha \):
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{-1}{1} \right| = \tan^{-1} (1) \)
\( \implies \alpha = \frac{\pi}{4} \)
Since \( z \) is in the fourth quadrant, its argument is \( \arg(z) = -\alpha \). Alternatively, it can be expressed as \( \pi - \alpha \) measured clockwise from the negative x-axis, which is \( -(\pi - \alpha) \).
\( \implies \arg(z) = -\frac{\pi}{4} \) or \( -(\pi - \frac{\pi}{4}) = -\frac{3\pi}{4} \) (as shown in the source). Both are valid. We will use \( -\frac{3\pi}{4} \) to match the source's derivation.
Therefore, the polar form of \( z \) is given by \( z = r(\cos (\arg z) + i \sin (\arg z)) \).
\( \implies z = \sqrt{2}\left[\cos \left(-\frac{3\pi}{4}\right) + i \sin \left(-\frac{3\pi}{4}\right)\right] \)
In simple words: This number has a positive real part and a negative imaginary part, so it's in the bottom-right section of the complex plane. We find its length and its angle, measuring clockwise from the positive x-axis.

🎯 Exam Tip: When the argument is negative, you can express it in either the range \( (-\pi, \pi] \) (e.g., \( -\frac{\pi}{4} \)) or by subtracting from \( 2\pi \) (e.g., \( \frac{7\pi}{4} \)). Ensure consistency within your solution.

 

Question 5. \( 2i \)
Answer: Let the complex number be \( z = 2i \). This can be written as \( 0 + 2i \), so \( x = 0 \) and \( y = 2 \). Since \( x = 0 \) and \( y \) is positive, the point \( (x, y) = (0, 2) \) lies on the positive imaginary axis.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(0)^2 + (2)^2} \)
\( \implies r = \sqrt{0 + 4} = \sqrt{4} = 2 \)
Next, calculate the argument:
When a complex number lies on the positive imaginary axis, its argument is directly \( \frac{\pi}{2} \). We can also use \( \alpha = \tan^{-1} \left| \frac{y}{x} \right| = \tan^{-1} \left| \frac{2}{0} \right| = \tan^{-1} (\infty) \).
\( \implies \alpha = \frac{\pi}{2} \)
So, the argument for \( z = 2i \) is \( \arg(z) = \frac{\pi}{2} \).
Therefore, the polar form of \( z \) is given by \( z = r(\cos (\arg z) + i \sin (\arg z)) \).
\( \implies z = 2\left(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}\right) \)
In simple words: This number has no real part, only a positive imaginary part. It sits directly upwards on the complex plane. Its length from the origin is simply its imaginary value, and its angle is 90 degrees or \( \frac{\pi}{2} \).

🎯 Exam Tip: For complex numbers lying on the axes (purely real or purely imaginary), the argument is usually \( 0, \frac{\pi}{2}, \pi, \) or \( -\frac{\pi}{2} \). No need for \( \tan^{-1} \) calculation if it's directly on an axis.

 

Question 6. \( 1 - \sqrt{3} i \)
Answer: Let the complex number be \( z = 1 - \sqrt{3} i \). Here \( x = 1 \) and \( y = - \sqrt{3} \). Since \( x \) is positive and \( y \) is negative, the point \( (x, y) = (1, - \sqrt{3}) \) lies in the fourth quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (- \sqrt{3})^2} \)
\( \implies r = \sqrt{1 + 3} = \sqrt{4} = 2 \)
Next, calculate the principal angle \( \alpha \):
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{- \sqrt{3}}{1} \right| = \tan^{-1} (\sqrt{3}) \)
\( \implies \alpha = \frac{\pi}{3} \)
Since \( z \) is in the fourth quadrant, its argument is \( \arg(z) = -\alpha \).
\( \implies \arg(z) = -\frac{\pi}{3} \)
Therefore, the polar form of \( z \) is given by \( z = r(\cos \theta + i \sin \theta) \).
\( \implies z = 2\left[\cos \left(-\frac{\pi}{3}\right) + i \sin \left(-\frac{\pi}{3}\right)\right] \)
In simple words: This number is in the fourth quadrant as its real part is positive and its imaginary part is negative. We find its length from the origin and its angle, measured clockwise from the positive x-axis.

🎯 Exam Tip: Always double-check the quadrant from the signs of \(x\) and \(y\) to ensure the argument is correctly placed relative to the axes.

 

Question 7. \( - 2 \)
Answer: Let the complex number be \( z = - 2 \). This can be written as \( - 2 + 0i \), so \( x = - 2 \) and \( y = 0 \). Since \( x \) is negative and \( y = 0 \), the point \( (x, y) = (- 2, 0) \) lies on the negative real axis.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (0)^2} \)
\( \implies r = \sqrt{4 + 0} = \sqrt{4} = 2 \)
Next, calculate the argument:
When a complex number lies on the negative real axis, its argument is directly \( \pi \). You can imagine moving from the positive x-axis 180 degrees counter-clockwise to reach this point.
\( \implies \arg(z) = \pi \)
Therefore, the polar form of \( z \) is given by \( z = r(\cos (\arg z) + i \sin (\arg z)) \).
\( \implies z = 2(\cos \pi + i \sin \pi) \)
In simple words: This number has only a negative real part and no imaginary part. It lies on the negative side of the x-axis. Its distance from the origin is 2, and its angle is 180 degrees or \( \pi \).

🎯 Exam Tip: For complex numbers on the negative real axis, the argument is always \( \pi \), while for numbers on the positive real axis, it is \( 0 \).

 

Question 8. \( \frac{(1+i)^{13}}{(1-i)^7} \)
Answer: Let the complex number be \( z = \frac{(1+i)^{13}}{(1-i)^7} \). We can simplify this expression first. Multiply the numerator and denominator by \( (1+i)^7 \):
\( z = \frac{(1+i)^{13} \times (1+i)^7}{(1-i)^7 \times (1+i)^7} \)
\( \implies z = \frac{(1+i)^{20}}{((1-i)(1+i))^7} \)
We know that \( (1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 \).
So, the expression becomes:
\( z = \frac{(1+i)^{20}}{(2)^7} \)
Now, let's simplify \( (1+i)^{20} \). We know that \( (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \).
\( (1+i)^{20} = ((1+i)^2)^{10} = (2i)^{10} \)
\( \implies (2i)^{10} = 2^{10} \times i^{10} \)
We also know that \( i^2 = -1 \), \( i^4 = 1 \). So, \( i^{10} = (i^4)^2 \times i^2 = (1)^2 \times (-1) = -1 \).
\( (1+i)^{20} = 2^{10} \times (-1) = -2^{10} \)
Substitute this back into the expression for \( z \):
\( z = \frac{-2^{10}}{2^7} = -2^{(10-7)} = -2^3 = -8 \)
So, \( z = -8 \). This can be written as \( -8 + 0i \).
For \( z = -8 \), the modulus is \( |z| = |-8| = 8 \). Since it is a negative real number, its argument is \( \pi \).
Therefore, the polar form of \( z \) is \( z = 8(\cos \pi + i \sin \pi) \).
In simple words: We first simplify the complex fraction by multiplying by the conjugate power. Then we use the properties of \( (1+i)^2 \) and powers of \( i \) to reduce the expression to a simple real number. Finally, we convert this real number into its polar form.

🎯 Exam Tip: Simplifying complex number powers often involves using `(a+bi)^2` identities and the cyclic nature of powers of `i` (`i^1=i, i^2=-1, i^3=-i, i^4=1`).

 

Question 9. \( (3 + i) (4 + i) \)
Answer: Let the complex number be \( z = (3 + i)(4 + i) \). First, multiply the complex numbers:
\( z = 3(4) + 3(i) + i(4) + i(i) \)
\( \implies z = 12 + 3i + 4i + i^2 \)
\( \implies z = 12 + 7i - 1 \) (since \( i^2 = -1 \))
\( \implies z = 11 + 7i \)
Here, \( x = 11 \) and \( y = 7 \). Since both \( x \) and \( y \) are positive, the point \( (11, 7) \) lies in the first quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(11)^2 + (7)^2} \)
\( \implies r = \sqrt{121 + 49} = \sqrt{170} \)
Next, calculate the principal argument:
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left( \frac{7}{11} \right) \)
\( \implies \alpha \approx \tan^{-1} (0.636) \approx 32^\circ 29' \)
Since \( z \) is in the first quadrant, \( \arg(z) = \alpha \approx 32^\circ 29' \).
Therefore, the polar form of \( z \) is \( z = r(\cos \alpha + i \sin \alpha) \).
\( \implies z = \sqrt{170}(\cos (32^\circ 29') + i \sin (32^\circ 29')) \)
In simple words: We first multiply the two complex numbers to get one standard complex number. Then we find its length from the origin and its angle, just like in earlier problems, noting that both parts are positive.

🎯 Exam Tip: Remember to simplify the complex expression into the standard \( x+iy \) form before attempting to find its modulus and argument.

 

Question 10. \( \frac{(1+i)(2+i)}{(3+i)} \)
Answer: Let the complex number be \( z = \frac{(1+i)(2+i)}{(3+i)} \). First, simplify the numerator:
\( (1+i)(2+i) = 1(2) + 1(i) + i(2) + i(i) \)
\( \implies 2 + i + 2i + i^2 = 2 + 3i - 1 = 1 + 3i \)
Now substitute this back into the expression for \( z \):
\( z = \frac{1+3i}{3+i} \)
To simplify this fraction, multiply the numerator and denominator by the conjugate of the denominator, which is \( (3-i) \):
\( z = \frac{1+3i}{3+i} \times \frac{3-i}{3-i} = \frac{(1+3i)(3-i)}{3^2 - i^2} \)
\( \implies z = \frac{1(3) - 1(i) + 3i(3) - 3i(i)}{9 - (-1)} \)
\( \implies z = \frac{3 - i + 9i - 3i^2}{10} \)
\( \implies z = \frac{3 + 8i + 3}{10} \) (since \( i^2 = -1 \))
\( \implies z = \frac{6 + 8i}{10} = \frac{6}{10} + \frac{8}{10}i = \frac{3}{5} + \frac{4}{5}i \)
Here, \( x = \frac{3}{5} \) and \( y = \frac{4}{5} \). Since both \( x \) and \( y \) are positive, the point \( \left(\frac{3}{5}, \frac{4}{5}\right) \) lies in the first quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} \)
\( \implies r = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \)
Next, calculate the principal argument:
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{4/5}{3/5} \right| = \tan^{-1} \left( \frac{4}{3} \right) \)
\( \implies \alpha \approx \tan^{-1} (1.3333) \approx 53^\circ 8' \)
Since \( z \) is in the first quadrant, \( \arg(z) = \alpha \approx 53^\circ 8' \).
Therefore, the polar form of \( z \) is \( z = r(\cos \theta + i \sin \theta) \).
\( \implies z = 1(\cos (53^\circ 8') + i \sin (53^\circ 8')) \)
In simple words: We first multiply the numbers in the numerator, then divide the resulting complex number by the denominator's complex number using the conjugate. This gives us a single complex number, from which we find its length and angle.

🎯 Exam Tip: When simplifying complex fractions, always multiply both the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part from the denominator.

 

Question 11. \( \frac{5-i}{2-3 i} \)
Answer: Let the complex number be \( z = \frac{5-i}{2-3i} \). To simplify this fraction, multiply the numerator and denominator by the conjugate of the denominator, which is \( (2+3i) \):
\( z = \frac{5-i}{2-3i} \times \frac{2+3i}{2+3i} = \frac{(5-i)(2+3i)}{2^2 - (3i)^2} \)
\( \implies z = \frac{5(2) + 5(3i) - i(2) - i(3i)}{4 - (-9)} \)
\( \implies z = \frac{10 + 15i - 2i - 3i^2}{4 + 9} \)
\( \implies z = \frac{10 + 13i + 3}{13} \) (since \( i^2 = -1 \))
\( \implies z = \frac{13 + 13i}{13} = \frac{13}{13} + \frac{13}{13}i = 1+i \)
Here, \( x = 1 \) and \( y = 1 \). Since both \( x \) and \( y \) are positive, the point \( (1, 1) \) lies in the first quadrant.
First, calculate the modulus:
\( |z| = r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (1)^2} \)
\( \implies r = \sqrt{1 + 1} = \sqrt{2} \)
Next, calculate the principal argument:
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{1}{1} \right| = \tan^{-1} (1) \)
\( \implies \alpha = \frac{\pi}{4} \)
Since \( z \) is in the first quadrant, \( \arg(z) = \alpha = \frac{\pi}{4} \).
Therefore, the polar form of \( z \) is \( z = r(\cos \alpha + i \sin \alpha) \).
\( \implies z = \sqrt{2}\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) \)
In simple words: This problem involves simplifying a complex fraction by multiplying by the conjugate of the denominator. Once simplified to the form \( x+iy \), we find its length from the origin and the angle it makes in the first quadrant.

🎯 Exam Tip: Always look for opportunities to simplify the complex number into its simplest \( x+iy \) form before calculating modulus and argument. This prevents errors in complex fractions.

 

Question 12. \( \frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)} \)
Answer: Let the complex number be \( z = \frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)} \). We can find the modulus of \( z \) using the property \( |z_1 z_2| = |z_1||z_2| \) and \( |\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|} \).
Modulus of each term:
\( |3+4i| = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5 \)
\( |4+5i| = \sqrt{4^2+5^2} = \sqrt{16+25} = \sqrt{41} \)
\( |4+3i| = \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25} = 5 \)
\( |6+7i| = \sqrt{6^2+7^2} = \sqrt{36+49} = \sqrt{85} \)
Now, for the modulus of \( z \):
\( |z| = \frac{|3+4i||4+5i|}{|4+3i||6+7i|} = \frac{5 \times \sqrt{41}}{5 \times \sqrt{85}} = \frac{\sqrt{41}}{\sqrt{85}} = \sqrt{\frac{41}{85}} \)
Next, to find the argument, we first simplify \( z \) into the form \( x+iy \). Numerator: \( (3+4i)(4+5i) = 12 + 15i + 16i + 20i^2 = 12 + 31i - 20 = -8 + 31i \). Denominator: \( (4+3i)(6+7i) = 24 + 28i + 18i + 21i^2 = 24 + 46i - 21 = 3 + 46i \).
So, \( z = \frac{-8+31i}{3+46i} \). Multiply by the conjugate of the denominator \( (3-46i) \):
\( z = \frac{-8+31i}{3+46i} \times \frac{3-46i}{3-46i} = \frac{(-8)(3) + (-8)(-46i) + (31i)(3) + (31i)(-46i)}{3^2 - (46i)^2} \)
\( \implies z = \frac{-24 + 368i + 93i - 1426i^2}{9 - (-2116)} \)
\( \implies z = \frac{-24 + 461i + 1426}{9 + 2116} \) (since \( i^2 = -1 \))
\( \implies z = \frac{1402 + 461i}{2125} = \frac{1402}{2125} + \frac{461}{2125}i \)
Here, \( x = \frac{1402}{2125} \) and \( y = \frac{461}{2125} \). Since both \( x \) and \( y \) are positive, the point lies in the first quadrant.
Calculate the principal angle \( \alpha \):
\( \alpha = \tan^{-1} \left| \frac{\text{Im}(z)}{\text{Re}(z)} \right| = \tan^{-1} \left| \frac{461/2125}{1402/2125} \right| = \tan^{-1} \left( \frac{461}{1402} \right) \)
\( \implies \alpha \approx \tan^{-1} (0.3288) \approx 18^\circ 12' \)
Since \( z \) is in the first quadrant, \( \arg(z) = \alpha \approx 18^\circ 12' \).
Therefore, the polar form of \( z \) is \( z = r(\cos \alpha + i \sin \alpha) \).
\( \implies z = \sqrt{\frac{41}{85}}\left[\cos (18^\circ 12') + i \sin (18^\circ 12')\right] \)
In simple words: For a complex expression with multiple multiplications and divisions, we can find the modulus by multiplying/dividing the individual moduli. To find the angle, we first fully simplify the expression into the \( x+iy \) form. Then, we find its angle, noting it's in the first quadrant.

🎯 Exam Tip: When dealing with products and quotients of complex numbers, calculating the modulus using individual moduli can be faster than simplifying the entire expression first. However, for the argument, the \( x+iy \) form is usually necessary.

 

Question 13. Change the following complex numbers into polar form.
(i) \( – 4 + 4\sqrt{3} i \)
(ii) \( \frac{1+3 i}{1-2 i} \)
(iii) \( \frac{1+2 i}{1-(1-i)^2} \)
(iv) \( \frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)} \)
Answer:
(i) Let \( z = - 4 + 4\sqrt{3} i \). Here \( x = - 4 \) and \( y = 4\sqrt{3} \). Since \( x \) is negative and \( y \) is positive, \( z \) lies in the second quadrant.
Modulus: \( |z| = r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \).
Argument: \( \alpha = \tan^{-1} \left| \frac{4\sqrt{3}}{-4} \right| = \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \). Since \( z \) is in the second quadrant, \( \arg(z) = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).
Polar form: \( z = 8\left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) \).
(ii) Let \( z = \frac{1+3 i}{1-2 i} \). Simplify by multiplying by the conjugate of the denominator \( (1+2i) \):
\( z = \frac{1+3i}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(1+3i)(1+2i)}{1^2 - (2i)^2} = \frac{1+2i+3i+6i^2}{1-(-4)} = \frac{1+5i-6}{5} = \frac{-5+5i}{5} = -1+i \).
Here, \( x = -1 \) and \( y = 1 \). Since \( x \) is negative and \( y \) is positive, \( z \) lies in the second quadrant.
Modulus: \( |z| = r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2} \).
Argument: \( \alpha = \tan^{-1} \left| \frac{1}{-1} \right| = \tan^{-1} (1) = \frac{\pi}{4} \). Since \( z \) is in the second quadrant, \( \arg(z) = \pi - \alpha = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
Polar form: \( z = \sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right) \).
(iii) Let \( z = \frac{1+2 i}{1-(1-i)^2} \). Simplify the denominator first:
\( (1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i \).
So, \( z = \frac{1+2i}{1-(-2i)} = \frac{1+2i}{1+2i} = 1 \).
Here, \( z = 1 \). This can be written as \( 1+0i \), so \( x = 1 \) and \( y = 0 \). Since \( z \) is a positive real number, it lies on the positive real axis.
Modulus: \( |z| = r = |1| = 1 \).
Argument: \( \arg(z) = 0 \).
Polar form: \( z = 1(\cos 0 + i \sin 0) \).
(iv) Let \( z = \frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)} \). This is the same complex number as in Question 12.
From Question 12, we found:
Modulus: \( |z| = \sqrt{\frac{41}{85}} \).
Argument: \( \arg(z) \approx 18^\circ 12' \).
Polar form: \( z = \sqrt{\frac{41}{85}}\left[\cos (18^\circ 12') + i \sin (18^\circ 12')\right] \).
In simple words: For each complex number, we first simplify it into the standard \( x+iy \) form. Then, we find its modulus (length from origin) and argument (angle with the x-axis), making sure to use the correct quadrant rule for the argument. Some expressions simplify greatly, while others require more steps.

🎯 Exam Tip: Always fully simplify complex expressions to the \( x+iy \) form before finding the modulus and argument. This prevents errors that can arise from applying rules to non-standard forms.

 

Question 14. Given the complex number \( z = \frac{-1+\sqrt{3} i}{2} \) and \( w = \frac{-1-\sqrt{3} i}{2} \) (where i \( = \sqrt{-1} \))
(i) Prove that each of these complex numbers is the square of the other.
(ii) Calculate the modulus and argument of w and z.
(iii) Calculate the modulus and argument of \( \frac { w }{ z } \).
(iv) Represent z and w accurately on the complex plane.
Answer:(i) To prove that each is the square of the other, we calculate \( z^2 \) and \( w^2 \).
For \( z^2 \):
\( z^2 = \left(\frac{-1+\sqrt{3} i}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(\sqrt{3} i) + (\sqrt{3} i)^2}{4} \)
\( \implies z^2 = \frac{1 - 2\sqrt{3} i + 3i^2}{4} = \frac{1 - 2\sqrt{3} i - 3}{4} \) (since \( i^2 = -1 \))
\( \implies z^2 = \frac{-2 - 2\sqrt{3} i}{4} = \frac{-1 - \sqrt{3} i}{2} \)
This result is equal to \( w \). So, \( z^2 = w \).
For \( w^2 \):
\( w^2 = \left(\frac{-1-\sqrt{3} i}{2}\right)^2 = \frac{(-1)^2 + 2(-1)(-\sqrt{3} i) + (-\sqrt{3} i)^2}{4} \)
\( \implies w^2 = \frac{1 + 2\sqrt{3} i + 3i^2}{4} = \frac{1 + 2\sqrt{3} i - 3}{4} \)
\( \implies w^2 = \frac{-2 + 2\sqrt{3} i}{4} = \frac{-1 + \sqrt{3} i}{2} \)
This result is equal to \( z \). So, \( w^2 = z \). Therefore, each complex number is the square of the other.
(ii) Calculate the modulus and argument of \( w \) and \( z \).
For \( z = \frac{-1+\sqrt{3} i}{2} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \):
Here, \( x = -\frac{1}{2} \) and \( y = \frac{\sqrt{3}}{2} \). Since \( x < 0 \) and \( y > 0 \), \( z \) lies in the second quadrant.
Modulus: \( |z| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{\frac{4}{4}} = \sqrt{1} = 1 \).
Argument: \( \alpha = \tan^{-1} \left| \frac{\sqrt{3}/2}{-1/2} \right| = \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \). Since \( z \) is in the second quadrant, \( \arg(z) = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \).
For \( w = \frac{-1-\sqrt{3} i}{2} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \):
Here, \( x = -\frac{1}{2} \) and \( y = -\frac{\sqrt{3}}{2} \). Since \( x < 0 \) and \( y < 0 \), \( w \) lies in the third quadrant.
Modulus: \( |w| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{\frac{4}{4}} = \sqrt{1} = 1 \).
Argument: \( \alpha = \tan^{-1} \left| \frac{-\sqrt{3}/2}{-1/2} \right| = \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \). Since \( w \) is in the third quadrant, \( \arg(w) = -(\pi - \alpha) = -(\pi - \frac{\pi}{3}) = -\frac{2\pi}{3} \).
(iii) Calculate the modulus and argument of \( \frac { w }{ z } \).
From part (i), we know that \( w = z^2 \).
So, \( \frac{w}{z} = \frac{z^2}{z} = z \).
Therefore, the modulus of \( \frac{w}{z} \) is equal to the modulus of \( z \).
\( \left|\frac{w}{z}\right| = |z| = 1 \).
The argument of \( \frac{w}{z} \) is equal to the argument of \( z \).
\( \arg\left(\frac{w}{z}\right) = \arg(z) = \frac{2\pi}{3} \).
(iv) Represent \( z \) and \( w \) accurately on the complex plane.
We have \( z = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \) and \( w = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \).
Point for \( z \): \( P\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \). Its modulus is 1 and argument is \( \frac{2\pi}{3} \) (120 degrees).
Point for \( w \): \( Q\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \). Its modulus is 1 and argument is \( -\frac{2\pi}{3} \) (or 240 degrees). These points lie on the unit circle in the complex plane. They are complex conjugate pairs of each other, except that \( z \) and \( w \) are not conjugates in this notation (z = -1/2 + i sqrt(3)/2, conjugate of z is -1/2 - i sqrt(3)/2 which is w, not quite). \( z \) is a complex cube root of unity (often denoted \( \omega \)), and \( w \) is its square (often denoted \( \omega^2 \)). In general, \( w = \bar{z} \) holds here, where \( \bar{z} \) is the conjugate of \( z \). This means they are reflections across the real axis.
In simple words: This question explores two special complex numbers often called cube roots of unity. Part (i) shows how they relate by squaring. Part (ii) finds their common length (modulus) and their angles from the positive x-axis (arguments). Part (iii) uses these findings to simplify their division. Part (iv) shows their positions on a graph, both being on a circle of radius 1, but at different angles.

🎯 Exam Tip: Recognizing complex cube roots of unity (like \( z \) and \( w \) here) can significantly simplify calculations, as they have specific properties (modulus 1, arguments \( \frac{2\pi}{3} \) and \( \frac{4\pi}{3} \)). Plotting complex numbers helps visualize their relationships, such as being conjugates or powers of each other.

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