Question 1 – Plot the following points

In this question, we have eight points given in polar coordinates (r, θ). Here r is the distance from the origin and θ is the angle from the positive x-axis.

For each point, we have to plot it on the Argand diagram. The first value is r and the second is θ.

Some r values are negative. When r is negative, it means we go in the opposite direction of the angle. So if r is negative, we add 180° to the angle and make r positive, then plot normally.

Some angles are in degrees, some are in radians, and some are negative. Negative angles mean we go clockwise instead of counterclockwise.

For each point, we first convert r to positive if needed, adjust the angle accordingly, then find the x and y coordinates using x = r cos θ and y = r sin θ. Then we plot the point (x, y) on the graph.

Question 2 – Express the following complex numbers in polar form

In this question, we have seven complex numbers. We have to convert each from rectangular form (x + iy) to polar form.

Polar form means writing the complex number as r(cos θ + i sin θ) or sometimes as r∠θ. Here r is the modulus and θ is the argument.

To find r, we use the formula r = √(x² + y²). This is the distance from the origin.

To find θ, we use tan θ = y/x. But we have to be careful about which quadrant the point lies in. For example, if x is negative and y is positive, θ is in the second quadrant, so we add 180° to the reference angle.

For parts (v), (vi), and (vii), we first simplify the fraction to get a single complex number in x + iy form. Then we find its polar form.

Question 3 – Convert from polar form to rectangular form

In this question, we are given complex numbers in polar form. Some are in trigonometric form r(cos θ + i sin θ). Some give only |z| and arg(z). We have to convert each to rectangular form x + iy.

To convert, we use the formulas:
x = r cos θ
y = r sin θ

For parts (iii), (iv), and (v), we are given |z| and arg(z) separately. So we directly plug into x = |z| cos(arg) and y = |z| sin(arg).

For part (vi), we have 2 cos(–33°) + i 2 sin(–33°). Here 2 is the modulus and –33° is the argument. So we find cos(–33°) = cos 33° and sin(–33°) = – sin 33°. Then we calculate the values.

Question 4 – Operations on polar form

In this question, we are given two complex numbers in polar form: z₁ = 9(cos(5π/4) + i sin(5π/4)) and z₂ = 5(cos(π/3) + i sin(π/3)).

We have to find four things:

• z₁ + z₂

• z₁ – z₂

• z₁ × z₂

• z₁ ÷ z₂

For addition and subtraction, we must first convert both numbers to rectangular form (x + iy). Then we add or subtract the real parts and the imaginary parts separately.

For multiplication, we use the polar form property: multiply the moduli and add the arguments. So |z₁ × z₂| = |z₁| × |z₂| and arg(z₁ × z₂) = arg(z₁) + arg(z₂).

For division, we use the property: divide the moduli and subtract the arguments. So |z₁ ÷ z₂| = |z₁| ÷ |z₂| and arg(z₁ ÷ z₂) = arg(z₁) – arg(z₂).

Question 5 – Operations on polar form (another example)

This question is exactly the same as question 4, but with different numbers. z₁ = 7(cos(23π/12) + i sin(23π/12)) and z₂ = 11(cos(11π/12) + i sin(11π/12)).

We do the same steps:

• Convert to rectangular form for addition and subtraction

• For multiplication, multiply moduli and add arguments

• For division, divide moduli and subtract arguments

After getting the answer in polar form, we convert it to x + iy form.

Question 6 – Prove properties of argument

In this question, we have to prove two important properties of arguments.

Part (i): Arg(z₁ × z₂) = Arg(z₁) + Arg(z₂)
This means when you multiply two complex numbers, their arguments add up.

Part (ii): Arg(z₁ ÷ z₂) = Arg(z₁) – Arg(z₂)
This means when you divide two complex numbers, the arguments subtract.

We prove these using the polar form of complex numbers. If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then when we multiply them, the angles add. This comes from trigonometric identities. Similarly for division, the angles subtract.

Question 7 – Divide two polar form numbers

In this question, we have z₁ = 6(cos 150° + i sin 150°) and z₂ = 3(cos 30° + i sin 30°). We have to divide z₁ by z₂.

We use the division property: divide the moduli and subtract the arguments.

So |z₁ ÷ z₂| = 6 ÷ 3 = 2.
arg(z₁ ÷ z₂) = 150° – 30° = 120°.

So the answer in polar form is 2(cos 120° + i sin 120°). Then we convert to rectangular form: cos 120° = –1/2 and sin 120° = √3/2. So the answer is 2(–1/2 + i√3/2) = –1 + i√3.

Question 8 – Multiply two polar form numbers

In this question, we have z₁ = 2(cos 60° + i sin 60°) and z₂ = 5(cos 90° + i sin 90°). We have to multiply them.

We use the multiplication property: multiply the moduli and add the arguments.

|z₁ × z₂| = 2 × 5 = 10.
arg(z₁ × z₂) = 60° + 90° = 150°.

So the answer in polar form is 10(cos 150° + i sin 150°). Then convert to rectangular form: cos 150° = –√3/2 and sin 150° = 1/2. So the answer is 10(–√3/2 + i/2) = –5√3 + 5i.

Question 9 – Find modulus and argument of z = –2 – 2i

We have the complex number z = –2 – 2i. Both the real part and the imaginary part are negative. So this point is in the third quadrant.

To find modulus: r = √(x² + y²) = √((–2)² + (–2)²) = √(4 + 4) = √8 = 2√2.

To find argument: tan θ = y/x = (–2)/(–2) = 1. The reference angle is 45°. But since the point is in the third quadrant, the argument is 180° + 45° = 225° or in radians, 5π/4.

So the modulus is 2√2 and the argument is 225° or 5π/4.

Question 10 – Write equation in cartesian form

We are given arg( \bar{z} – 2 + i ) = 2π/3, where z = x + iy.

First, \bar{z} = x – iy. Then \bar{z} – 2 + i = (x – 2) + i(–y + 1) = (x – 2) + i(1 – y).

The argument of this complex number is 2π/3 (which is 120°). For an angle of 120°, the tangent is negative. Also, tan θ = y/x.

So we write tan(2π/3) = (1 – y)/(x – 2). tan(2π/3) = –√3.

So (1 – y)/(x – 2) = –√3. Cross multiply: 1 – y = –√3(x – 2) = –√3x + 2√3.

Then bring terms together to get the equation of a straight line.

Question 11 – Show a circle equation

We are given that arg( (\bar{z} – 1 + 2i) / (\bar{z} + 1 – 2i) ) = 9π/4.

First, we simplify. 9π/4 is more than 2π. 9π/4 – 2π = 9π/4 – 8π/4 = π/4. So the argument is π/4 (45°).

Now we use the property that the argument of a fraction is the difference of arguments. So arg( numerator ) – arg( denominator ) = π/4.

Then we put z = x + iy and \bar{z} = x – iy. The numerator is (x – 1) + i(–y + 2). The denominator is (x + 1) + i(–y – 2).

When two arguments have a fixed difference, it means the points form a circle. By squaring and simplifying, we get the equation x² + y² – 4x + 2y – 5 = 0, which is a circle.

Question 12 – Show a linear equation

We are given arg(z – 2 – 3i) – arg(z + 2 + 3i) = 2π.

If the difference of two arguments is 2π, it means the two vectors point in the same direction. That means (z – 2 – 3i) and (z + 2 + 3i) are collinear and point the same way.

This happens when the ratio (z – 2 – 3i) / (z + 2 + 3i) is a positive real number. That means its imaginary part is zero.

When we set the imaginary part to zero and simplify, we get the equation 2y = 3x, which is a straight line.

Question 13 – Solve |z – 2i| = |\bar{z} + 2|

We have z = x + iy. Then \bar{z} = x – iy.

First, find |z – 2i|. z – 2i = x + iy – 2i = x + i(y – 2). Its modulus is √(x² + (y – 2)²).

Next, find |\bar{z} + 2|. \bar{z} + 2 = (x – iy) + 2 = (x + 2) – iy. Its modulus is √((x + 2)² + (–y)²) = √((x + 2)² + y²).

Set them equal: √(x² + (y – 2)²) = √((x + 2)² + y²).

Square both sides: x² + (y – 2)² = (x + 2)² + y².

Expand: x² + y² – 4y + 4 = x² + 4x + 4 + y².

Cancel x² and y² from both sides. Then –4y + 4 = 4x + 4. Cancel 4 from both sides: –4y = 4x. So y = –x.

So the solution is all points on the line y = –x.

Question 14 – Solve another modulus equation

We have |5z + 4 + i| = |5\bar{z} – 3 + 2i|.

Put z = x + iy and \bar{z} = x – iy.

First, 5z + 4 + i = 5x + 5iy + 4 + i = (5x + 4) + i(5y + 1).

Its modulus squared is (5x + 4)² + (5y + 1)².

Next, 5\bar{z} – 3 + 2i = 5x – 5iy – 3 + 2i = (5x – 3) + i(–5y + 2).

Its modulus squared is (5x – 3)² + (–5y + 2)².

Set them equal, expand both sides, simplify. The square terms will cancel, and we will get a linear equation in x and y. That line is the solution.

Question 15 – Determine the set of points

We have |3\bar{z} – 2 + i| = |3z + i|.

Again put z = x + iy and \bar{z} = x – iy.

First, 3\bar{z} – 2 + i = 3x – 3iy – 2 + i = (3x – 2) + i(–3y + 1).

Its modulus squared is (3x – 2)² + (–3y + 1)².

Next, 3z + i = 3x + 3iy + i = 3x + i(3y + 1).

Its modulus squared is (3x)² + (3y + 1)².

Set them equal, expand, and simplify. The x² and y² terms will cancel. We will get a linear equation. The solution is all points on that line.

Question 16 – Show that |w| = 1 implies z is real

We have w = (1 – iz) / (z – i) and z = x + iy. We are given that |w| = 1. We have to show that z is real, meaning y = 0.

We put z = x + iy into the expression. Then simplify the fraction. After simplifying, we find |w|². Set |w|² = 1.

When we do this, the imaginary terms will cancel and we will be left with an equation that forces y = 0. This means z has no imaginary part, so z is purely real.

Question 17 – Find the expression

We have two complex numbers z₁ and z₂, with |z₂| = 1. We have to find |(z₂ – z₁) / (1 – \bar{z₁} z₂)|.

This is a standard result. The numerator is z₂ – z₁. The denominator is 1 – \bar{z₁} z₂.

We use the property that |1 – \bar{z₁} z₂| = |z₂ – z₁| when |z₂| = 1. This is because multiplying by something of modulus 1 does not change the modulus.

So the whole fraction has modulus equal to 1. Therefore the answer is 1.

Question 18 – AC circuit application (current in polar form)

We have voltage V = 120(cos(π/4) + i sin(π/4)) volts. Impedance Z = (1 + i√3)/2 ohms.

We need to find current I. In AC circuits, Ohm’s law works with complex numbers: V = I × Z, so I = V / Z.

We divide V by Z. V is already in polar form: modulus 120, argument π/4. We convert Z to polar form first. Z = (1 + i√3)/2. Its modulus is √(1² + (√3)²)/2 = √4/2 = 2/2 = 1. Its argument is tan⁻¹(√3/1) = 60° = π/3.

So I = 120∠(π/4) ÷ 1∠(π/3) = 120∠(π/4 – π/3) = 120∠(3π/12 – 4π/12) = 120∠(–π/12).

So the current in polar form is I = 120(cos(–π/12) + i sin(–π/12)).

Question 19 – AC circuit application (current in both forms)

We have voltage V = 90 + 30i volts and impedance Z = 3 – 6i ohms. We need to find current I = V / Z.

First, we divide using the conjugate of the denominator. Multiply numerator and denominator by the conjugate of Z, which is 3 + 6i.

So I = (90 + 30i) / (3 – 6i) = (90 + 30i)(3 + 6i) / ((3 – 6i)(3 + 6i)).

The denominator is 3² + 6² = 9 + 36 = 45.

The numerator: (90 + 30i)(3 + 6i) = 90×3 + 90×6i + 30i×3 + 30i×6i = 270 + 540i + 90i + 180i². Since i² = –1, 180i² = –180. So real part: 270 – 180 = 90. Imaginary part: 540i + 90i = 630i. So numerator = 90 + 630i.

So I = (90 + 630i) / 45 = 90/45 + (630/45)i = 2 + 14i. This is rectangular form.

To get polar form, find modulus: √(2² + 14²) = √(4 + 196) = √200 = 10√2. Argument: tan⁻¹(14/2) = tan⁻¹(7). So the polar form is 10√2(cos θ + i sin θ) where θ = tan⁻¹(7).

Question 20 – Encrypt the word “CODE”

In this question, we use complex numbers for encryption. Each letter is first converted to a number. Usually A=1, B=2, C=3, etc. Or sometimes A=0, B=1, etc. We need to use the given mapping.

The word “CODE” has four letters. We treat each letter as a real number. Then we multiply each number by the encryption key k = 2 – i. This gives us a complex number for each letter. That is the encrypted form.

To decrypt, we multiply by the inverse of the key. The inverse of (2 – i) is 1/(2 – i) = (2 + i)/(4 + 1) = (2 + i)/5. So we multiply the encrypted numbers by (2 + i)/5 to get back the original numbers, then convert back to letters.

Question 21 – Encrypt the word “QUIZ”

This is similar to question 20. The encryption key is k = 3 – 3i. First, convert each letter of “QUIZ” to a number using the given mapping (likely A=1, B=2, etc.). Then multiply each number by (3 – 3i) to get the encrypted complex numbers.

To decrypt, we find the inverse of the key. 1/(3 – 3i) = (3 + 3i)/(9 + 9) = (3 + 3i)/18 = (1 + i)/6. Then multiply each encrypted number by (1 + i)/6 to recover the original numbers, then convert back to letters.

Question 22 – Encrypt the word “CLASS” by addition

This question is different. Here we encrypt by adding a key, not multiplying. The key is k = –3 + 4i.

First, convert each letter of “CLASS” to a number. Then add the complex number (–3 + 4i) to each number. So each letter becomes a complex number: (letter number – 3) + i(4).

To decrypt, we subtract the key. So we take the encrypted complex number and subtract (–3 + 4i). That means we add 3 to the real part and subtract 4 from the imaginary part. Then we get back the original numbers, and we convert back to letters.