Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.5

Question 1. Find the modulus of the following complex numbers.
(i) \( \frac{2}{3+4i} \)
(ii) \( \frac{2-i}{1+i}+\frac{1-2 i}{1-i} \)
(iii) \( |(1 – i)^{10}| \)
(iv) \( |2i(3 – 4i) (4 – 3i)| \)
Answer:
(i) Let \( z = \frac{2i}{3+4i} \). To simplify this complex number, we multiply the numerator and denominator by the conjugate of the denominator, which is \( 3-4i \).
\( \implies z = \frac{2i}{3+4i} \times \frac{3-4i}{3-4i} \)
\( \implies z = \frac{6i - 8i^2}{3^2 - (4i)^2} \)
\( \implies z = \frac{6i + 8}{9 + 16} \)
\( \implies z = \frac{8+6i}{25} \)
The modulus of \( z \) is \( |z| = \sqrt{\left(\frac{8}{25}\right)^2 + \left(\frac{6}{25}\right)^2} \)
\( \implies |z| = \frac{1}{25} \sqrt{8^2 + 6^2} \)
\( \implies |z| = \frac{1}{25} \sqrt{64 + 36} \)
\( \implies |z| = \frac{1}{25} \sqrt{100} \)
\( \implies |z| = \frac{10}{25} \)
\( \implies |z| = \frac{2}{5} \)
(ii) Let \( z = \frac{2-i}{1+i}+\frac{1-2i}{1-i} \). To combine these fractions, we find a common denominator and add them.
\( \implies z = \frac{(2-i)(1-i) + (1-2i)(1+i)}{(1+i)(1-i)} \)
\( \implies z = \frac{(2 - 2i - i + i^2) + (1 + i - 2i - 2i^2)}{1 - i^2} \)
\( \implies z = \frac{(2 - 3i - 1) + (1 - i + 2)}{1 + 1} \)
\( \implies z = \frac{(1 - 3i) + (3 - i)}{2} \)
\( \implies z = \frac{4 - 4i}{2} \)
\( \implies z = 2 - 2i \)
The modulus of \( z \) is \( |z| = \sqrt{2^2 + (-2)^2} \)
\( \implies |z| = \sqrt{4 + 4} \)
\( \implies |z| = \sqrt{8} \)
\( \implies |z| = 2\sqrt{2} \)
(iii) To find the modulus of a complex number raised to a power, we can first find the modulus of the base complex number and then raise that result to the given power.
\( |(1 – i)^{10}| = (|1 – i|)^{10} \)
First, find \( |1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2} \).
\( \implies |(1 – i)^{10}| = (\sqrt{2})^{10} \)
\( \implies |(1 – i)^{10}| = (2^{\frac{1}{2}})^{10} \)
\( \implies |(1 – i)^{10}| = 2^5 \)
\( \implies |(1 – i)^{10}| = 32 \)
(iv) When finding the modulus of a product of complex numbers, we can find the modulus of each individual complex number and then multiply those moduli together. This property simplifies the calculation significantly.
\( |2i(3 – 4i) (4 – 3i)| = |2i| \times |3 – 4i| \times |4 – 3i| \)
First, find the individual moduli:
\( |2i| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2 \)
\( |3-4i| = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5 \)
\( |4-3i| = \sqrt{4^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5 \)
Now, multiply these moduli:
\( \implies |2i(3 – 4i) (4 – 3i)| = 2 \times 5 \times 5 \)
\( \implies |2i(3 – 4i) (4 – 3i)| = 50 \)In simple words: For each complex number expression, we simplify it into the form \( a+bi \) and then find its modulus using the formula \( \sqrt{a^2+b^2} \). For powers and products, we can use properties like \( |z^n| = |z|^n \) and \( |z_1 z_2| = |z_1| |z_2| \) to make the calculations easier.

🎯 Exam Tip: Remember that \( |z_1 z_2| = |z_1| |z_2| \) and \( |\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|} \) are very useful properties for simplifying modulus calculations involving products and quotients.

Question 2. For any two complex numbers \( z_1 \) and \( z_2 \), such that \( |z_1| = |z_2| = 1 \) and \( z_1 z_2 \neq -1 \), then show that \( \frac{z_{1}+z_{2}}{1+z_{1} z_{2}} \) is a real number.
Answer: To show that a complex number \( z \) is a real number, we must prove that it is equal to its complex conjugate, i.e., \( z = \bar{z} \).
We are given \( |z_1| = 1 \) and \( |z_2| = 1 \).
From the property \( |z|^2 = z\bar{z} \), we have:
\( |z_1|^2 = z_1\bar{z}_1 = 1 \implies \bar{z}_1 = \frac{1}{z_1} \)
\( |z_2|^2 = z_2\bar{z}_2 = 1 \implies \bar{z}_2 = \frac{1}{z_2} \)
Let \( z = \frac{z_{1}+z_{2}}{1+z_{1} z_{2}} \). Now, let's find the conjugate of \( z \):
\( \bar{z} = \overline{\left(\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}\right)} \)
\( \implies \bar{z} = \frac{\overline{z_1}+\overline{z_2}}{\overline{1}+\overline{z_1}\overline{z_2}} \) (Using conjugate properties: \( \overline{z_1+z_2} = \overline{z_1}+\overline{z_2} \) and \( \overline{z_1 z_2} = \overline{z_1}\overline{z_2} \))
\( \implies \bar{z} = \frac{\bar{z}_1+\bar{z}_2}{1+\bar{z}_1\bar{z}_2} \)
Now substitute \( \bar{z}_1 = \frac{1}{z_1} \) and \( \bar{z}_2 = \frac{1}{z_2} \):
\( \implies \bar{z} = \frac{\frac{1}{z_1}+\frac{1}{z_2}}{1+\frac{1}{z_1}\frac{1}{z_2}} \)
\( \implies \bar{z} = \frac{\frac{z_2+z_1}{z_1 z_2}}{\frac{z_1 z_2+1}{z_1 z_2}} \)
\( \implies \bar{z} = \frac{z_1+z_2}{1+z_1 z_2} \)
Since \( \bar{z} = z \), the given complex number \( z \) is a real number. This property is often useful in various complex number proofs.In simple words: To show the complex number is real, we check if it is the same as its conjugate. Using the given condition that the moduli are 1, we can write the conjugate of each \( z \) as \( \frac{1}{z} \). When we put these into the conjugate of the whole expression, we find it simplifies back to the original expression, which proves it is a real number.

🎯 Exam Tip: The key idea for proving a complex number is real is to show \( z = \bar{z} \). Remember the property \( z\bar{z} = |z|^2 \) and how to use it when \( |z|=1 \).

Question 3. Which one of the points \( 10 – 8i, 11 + 6i \) is closest to \( 1 + i \).
Answer: To find which point is closest to \( 1+i \), we need to calculate the distance between \( 1+i \) and each of the other two points. The distance between two complex numbers \( z_1 \) and \( z_2 \) is given by the modulus \( |z_1 - z_2| \). Let the point \( 1+i \) be A, \( 10-8i \) be B, and \( 11+6i \) be C.
First, calculate the distance between A and B, which is \( |AB| \):
\( |AB| = |(10 – 8i) – (1 + i)| \)
\( \implies |AB| = |10 – 8i – 1 – i| \)
\( \implies |AB| = |9 – 9i| \)
\( \implies |AB| = \sqrt{9^2 + (-9)^2} \)
\( \implies |AB| = \sqrt{81 + 81} \)
\( \implies |AB| = \sqrt{162} \)
\( \implies |AB| = 9\sqrt{2} \approx 9 \times 1.414 = 12.726 \)
Next, calculate the distance between A and C, which is \( |CA| \):
\( |CA| = |(11 + 6i) – (1 + i)| \)
\( \implies |CA| = |11 + 6i - 1 - i| \)
\( \implies |CA| = |10 + 5i| \)
\( \implies |CA| = \sqrt{10^2 + 5^2} \)
\( \implies |CA| = \sqrt{100 + 25} \)
\( \implies |CA| = \sqrt{125} \)
\( \implies |CA| = 5\sqrt{5} \approx 5 \times 2.236 = 11.18 \)
Comparing the distances, \( 11.18 \) is less than \( 12.726 \).
Therefore, the point \( 11+6i \) is closest to \( 1+i \). This method uses the geometric interpretation of complex numbers to find distances.In simple words: To find the closest point, we measure the distance from \( 1+i \) to each of the other two points. We use the absolute value formula \( |z_1 - z_2| \) for distance. The point that gives a smaller distance is the one that is closest. Here, \( 11+6i \) is closer than \( 10-8i \).

🎯 Exam Tip: Remember that the distance between two complex numbers \( z_1 \) and \( z_2 \) is simply \( |z_1 - z_2| \). It's similar to finding the distance between two points in a 2D plane.

Question 4. If \( |z| = 3 \), show that \( 7 \le |z + 6 – 8i| \le 13 \).
Answer: To show this inequality, we use the triangle inequality theorem for complex numbers. This theorem states that for any two complex numbers \( z_1 \) and \( z_2 \): \( ||z_1| - |z_2|| \le |z_1 + z_2| \le |z_1| + |z_2| \).
Let \( z_1 = z \) and \( z_2 = 6-8i \). We are given that \( |z| = 3 \).
First, we calculate the modulus of \( z_2 \):
\( |6-8i| = \sqrt{6^2 + (-8)^2} \)
\( \implies |6-8i| = \sqrt{36 + 64} \)
\( \implies |6-8i| = \sqrt{100} \)
\( \implies |6-8i| = 10 \)
Now, we apply the upper bound of the triangle inequality: \( |z_1 + z_2| \le |z_1| + |z_2| \)
\( |z + 6 – 8i| \le |z| + |6 – 8i| \)
\( \implies |z + 6 – 8i| \le 3 + 10 \)
\( \implies |z + 6 – 8i| \le 13 \) .......... (1)
Next, we apply the lower bound of the triangle inequality: \( |z_1 + z_2| \ge ||z_1| - |z_2|| \)
\( |z + 6 – 8i| \ge ||z| - |6 – 8i|| \)
\( \implies |z + 6 – 8i| \ge |3 - 10| \)
\( \implies |z + 6 – 8i| \ge |-7| \)
\( \implies |z + 6 – 8i| \ge 7 \) .......... (2)
Combining results from (1) and (2), we get:
\( 7 \le |z + 6 – 8i| \le 13 \)
Hence, the inequality is proved. This method is fundamental for understanding complex number bounds.In simple words: We use a special rule called the triangle inequality for complex numbers. It tells us that the length of the sum of two complex numbers is between the difference and sum of their individual lengths. We calculate the length of \( 6-8i \), which is 10. Then, using the rule with \( |z|=3 \) and \( |6-8i|=10 \), we find that the total length must be between \( |3-10| \) and \( 3+10 \), which means it's between 7 and 13.

🎯 Exam Tip: Clearly state the triangle inequality you are using. Remember to take the absolute value for the lower bound: \( ||z_1| - |z_2|| \). This ensures the lower bound is always positive.

Question 5. If \( |z| = 1 \), show that \( 2 \le |z² – 3| \le 4 \).
Answer: To demonstrate this inequality, we use the triangle inequality theorem for complex numbers, which states that \( ||z_1| - |z_2|| \le |z_1 + z_2| \le |z_1| + |z_2| \).
Let \( z_1 = z^2 \) and \( z_2 = -3 \).
We are given \( |z| = 1 \).
First, calculate the modulus of \( z_1 \) and \( z_2 \):
\( |z_1| = |z^2| = |z|^2 = 1^2 = 1 \)
\( |z_2| = |-3| = 3 \)
Now, apply the upper bound of the triangle inequality: \( |z_1 + z_2| \le |z_1| + |z_2| \)
\( |z^2 – 3| \le |z^2| + |-3| \)
\( \implies |z^2 – 3| \le 1 + 3 \)
\( \implies |z^2 – 3| \le 4 \)
Next, apply the lower bound of the triangle inequality: \( |z_1 + z_2| \ge ||z_1| - |z_2|| \)
\( |z^2 – 3| \ge ||z^2| - |-3|| \)
\( \implies |z^2 – 3| \ge |1 - 3| \)
\( \implies |z^2 – 3| \ge |-2| \)
\( \implies |z^2 – 3| \ge 2 \)
Combining these results, we get:
\( 2 \le |z^2 – 3| \le 4 \)
Hence, the inequality is proved. This shows how to establish bounds for the modulus of a complex expression.In simple words: We use the triangle inequality rule for complex numbers. We know \( |z|=1 \), so \( |z^2| \) is also 1. We also know that \( |-3| \) is 3. The rule says that \( |z^2-3| \) must be between \( |1-3| \) and \( 1+3 \). This means it is between \( |-2| \) (which is 2) and 4. So the answer is 2 to 4.

🎯 Exam Tip: Remember that \( |z^n| = |z|^n \). This is crucial when dealing with powers of complex numbers in inequalities and modulus calculations.

Question 6. If \( |z| = 2 \) show that the \( 8 \le |z + 6 + 8i| \le 12 \).
Answer: To prove this inequality, we use the triangle inequality theorem for complex numbers, which states that \( ||z_1| - |z_2|| \le |z_1 + z_2| \le |z_1| + |z_2| \).
Let \( z_1 = z \) and \( z_2 = 6+8i \). We are given that \( |z| = 2 \).
First, we calculate the modulus of \( z_2 \):
\( |6+8i| = \sqrt{6^2 + 8^2} \)
\( \implies |6+8i| = \sqrt{36 + 64} \)
\( \implies |6+8i| = \sqrt{100} \)
\( \implies |6+8i| = 10 \)
Now, we apply the upper bound of the triangle inequality: \( |z_1 + z_2| \le |z_1| + |z_2| \)
\( |z + 6 + 8i| \le |z| + |6 + 8i| \)
\( \implies |z + 6 + 8i| \le 2 + 10 \)
\( \implies |z + 6 + 8i| \le 12 \) .......... (1)
Next, we apply the lower bound of the triangle inequality: \( |z_1 + z_2| \ge ||z_1| - |z_2|| \)
\( |z + 6 + 8i| \ge ||z| - |6 + 8i|| \)
\( \implies |z + 6 + 8i| \ge |2 - 10| \)
\( \implies |z + 6 + 8i| \ge |-8| \)
\( \implies |z + 6 + 8i| \ge 8 \) .......... (2)
Combining results from (1) and (2), we get:
\( 8 \le |z + 6 + 8i| \le 12 \)
Hence, the inequality is proved. This demonstrates how to determine the range of possible moduli for a complex expression.In simple words: We use the triangle inequality, which sets limits for the length of a complex number sum. We know \( |z|=2 \), and we calculate the length of \( 6+8i \) to be 10. The rule says the length of \( |z+6+8i| \) is between \( |2-10| \) (which is 8) and \( 2+10 \) (which is 12). So it is between 8 and 12.

🎯 Exam Tip: Always correctly identify \( z_1 \) and \( z_2 \) in the expression. Pay attention to the signs when calculating the lower bound to ensure you get a positive absolute difference.

Question 7. If \( z_1, z_2 \) and \( z_3 \) are three complex numbers such that \( |z_1| = 1, |z_2| = 2, |z_3| = 3 \) and \( |z_1 + z_2 + z_3| = 1 \) show that \( |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| = 6 \).
Answer: To show the given equality, we start by manipulating the expression \( |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| \).
We can factor out \( z_1 z_2 z_3 \) from the expression:
\( |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| = |z_1 z_2 z_3 \left( \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right)| \)
Using the property \( |w_1 w_2| = |w_1| |w_2| \):
\( \implies |z_1| |z_2| |z_3| \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} \right| \)
We know the property \( |z|^2 = z\bar{z} \). From this, we can write \( \bar{z} = \frac{|z|^2}{z} \).
Using the given moduli:
For \( z_1 \): \( |z_1| = 1 \implies \bar{z}_1 = \frac{|z_1|^2}{z_1} = \frac{1^2}{z_1} = \frac{1}{z_1} \)
For \( z_2 \): \( |z_2| = 2 \implies \bar{z}_2 = \frac{|z_2|^2}{z_2} = \frac{2^2}{z_2} = \frac{4}{z_2} \)
For \( z_3 \): \( |z_3| = 3 \implies \bar{z}_3 = \frac{|z_3|^2}{z_3} = \frac{3^2}{z_3} = \frac{9}{z_3} \)
Now, substitute these into the expression:
\( = |z_1| |z_2| |z_3| |\bar{z}_3 + \bar{z}_2 + \bar{z}_1| \)
Using the property \( \overline{w_1+w_2+w_3} = \bar{w}_1+\bar{w}_2+\bar{w}_3 \):
\( = |z_1| |z_2| |z_3| |\overline{z_1 + z_2 + z_3}| \)
Since \( |\bar{w}| = |w| \):
\( = |z_1| |z_2| |z_3| |z_1 + z_2 + z_3| \)
Finally, substitute the given values of the moduli: \( |z_1|=1, |z_2|=2, |z_3|=3 \) and \( |z_1+z_2+z_3|=1 \).
\( = (1)(2)(3)(1) \)
\( = 6 \)
Thus, \( |9z_1 z_2 + 4z_1 z_3 + z_2 z_3| = 6 \). The careful use of modulus properties helps simplify complex expressions.In simple words: We start with the complex expression and take out the product \( z_1 z_2 z_3 \). Then, we use a rule that connects the conjugate of a complex number to its modulus. By substituting the conjugates based on the given moduli, the expression inside the absolute value becomes the conjugate of \( z_1+z_2+z_3 \). Since the length of a complex number is the same as its conjugate, we can replace it with \( |z_1+z_2+z_3| \). Finally, we multiply all the given lengths to get the answer, which is 6.

🎯 Exam Tip: Remember the critical property \( z\bar{z} = |z|^2 \), which implies \( \bar{z} = \frac{|z|^2}{z} \). This substitution is key when you have inverse terms like \( \frac{1}{z} \) and known moduli.

Question 8. If the area of the triangle formed by the vertices \( z, iz \), and \( z + iz \) is 50 square units, find the value of \( |z| \).
Answer: The vertices of the triangle are \( z, iz, \) and \( z+iz \).
Let's consider the vectors representing the sides of the triangle from one vertex. If we consider `\( z \)` as one vertex, the other two vectors from `\( z \)` are `\( iz - z \)` and `\( (z+iz) - z = iz \)`.
The sides of the triangle are represented by the complex numbers:
Side 1: `\( z \)` (from origin to `\( z \)` is `\( z \)` itself, in relative terms this could be length of `\( z \)`)
Side 2: `\( iz \)` (from origin to `\( iz \)`)
Side 3: `\( (z+iz) \)` (from origin to `\( z+iz \)`)
The vectors corresponding to the vertices `\( z \)` and `\( iz \)` from the origin are perpendicular. This is because multiplying a complex number by \( i \) rotates it by 90 degrees counter-clockwise.
So, the triangle formed by the origin, \( z \), and \( iz \) is a right-angled triangle.
The lengths of the legs of this right-angled triangle are \( |z| \) and \( |iz| \).
We know that \( |iz| = |i| |z| = 1 \times |z| = |z| \).
Therefore, the triangle has legs of length \( |z| \) and \( |z| \). The area of a right-angled triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).
Area \( = \frac{1}{2} \times |z| \times |z| \)
Area \( = \frac{1}{2} |z|^2 \)
We are given that the area is 50 square units.
\( \frac{1}{2} |z|^2 = 50 \)
\( \implies |z|^2 = 100 \)
\( \implies |z| = \sqrt{100} \)
\( \implies |z| = 10 \)
The value of \( |z| \) is 10. Understanding the geometric representation of complex numbers helps solve such problems.In simple words: The three points \( z, iz, \) and \( z+iz \) form a special kind of triangle. The points \( z \) and \( iz \) are at right angles to each other if we think of them starting from zero. This means the triangle is a right-angled triangle with two sides of length \( |z| \). The area of this triangle is \( \frac{1}{2} \times |z| \times |z| \). We are told the area is 50, so we set up the equation \( \frac{1}{2} |z|^2 = 50 \) and solve for \( |z| \), which gives 10.

🎯 Exam Tip: Recognize that multiplying a complex number by \( i \) rotates it by 90 degrees. This geometric insight simplifies many problems involving complex numbers and shapes.

Question 9. Show that the equation \( z^3 + 2 \bar {z} = 0 \) has five solutions.
Answer: To show that the equation \( z^3 + 2 \bar {z} = 0 \) has five solutions, we first rearrange the equation:
\( z^3 = -2 \bar {z} \)
Take the modulus on both sides:
\( |z^3| = |-2 \bar {z}| \)
Using the properties \( |z^n| = |z|^n \), \( |c z| = |c| |z| \) and \( |\bar {z}| = |z| \):
\( |z|^3 = |-2| |\bar {z}| \)
\( \implies |z|^3 = 2|z| \)
Rearrange this equation:
\( |z|^3 - 2|z| = 0 \)
Factor out \( |z| \):
\( |z| (|z|^2 - 2) = 0 \)
This gives two possible cases for \( |z| \):
**Case 1: \( |z| = 0 \)**
If \( |z| = 0 \), then \( z = 0 \). This is one solution.
**Case 2: \( |z|^2 - 2 = 0 \)**
If \( |z|^2 - 2 = 0 \), then \( |z|^2 = 2 \).
We know that \( z\bar{z} = |z|^2 \), so \( z\bar{z} = 2 \).
This implies \( \bar{z} = \frac{2}{z} \).
Substitute \( \bar{z} = \frac{2}{z} \) back into the original equation \( z^3 = -2 \bar{z} \):
\( z^3 = -2 \left( \frac{2}{z} \right) \)
\( \implies z^3 = -\frac{4}{z} \)
Multiply both sides by \( z \) (since \( z \neq 0 \) in this case):
\( z^4 = -4 \)
To find the solutions for \( z^4 = -4 \), we express \( -4 \) in polar form.
\( -4 = 4 (\cos(\pi + 2k\pi) + i \sin(\pi + 2k\pi)) \), where \( k \) is an integer.
The fourth roots are:
\( z = (4)^{\frac{1}{4}} \left( \cos\left(\frac{\pi + 2k\pi}{4}\right) + i \sin\left(\frac{\pi + 2k\pi}{4}\right) \right) \) for \( k = 0, 1, 2, 3 \).
\( \implies z = \sqrt{2} \left( \cos\left(\frac{\pi + 2k\pi}{4}\right) + i \sin\left(\frac{\pi + 2k\pi}{4}\right) \right) \)
For \( k=0 \): \( z_0 = \sqrt{2} (\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})) = \sqrt{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = 1+i \)
For \( k=1 \): \( z_1 = \sqrt{2} (\cos(\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4})) = \sqrt{2} (-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = -1+i \)
For \( k=2 \): \( z_2 = \sqrt{2} (\cos(\frac{5\pi}{4}) + i \sin(\frac{5\pi}{4})) = \sqrt{2} (-\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}) = -1-i \)
For \( k=3 \): \( z_3 = \sqrt{2} (\cos(\frac{7\pi}{4}) + i \sin(\frac{7\pi}{4})) = \sqrt{2} (\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}) = 1-i \)
These are 4 distinct non-zero solutions.
Including the solution from Case 1 (\( z=0 \)), the total number of solutions is \( 1 + 4 = 5 \). This comprehensive approach demonstrates the power of complex number theory.In simple words: We start by rearranging the equation and taking the 'length' (modulus) of both sides. This helps us find two main possibilities for the length of \( z \): either it's 0 (meaning \( z=0 \), one solution) or its square is 2. If its square is 2, we replace \( \bar{z} \) in the original equation and get \( z^4 = -4 \). We then find the four roots of this equation using polar form. Adding these four roots to the \( z=0 \) solution gives a total of five solutions.

🎯 Exam Tip: When solving equations involving both \( z \) and \( \bar{z} \), taking the modulus is often the first step to find the magnitude of possible solutions. Don't forget the \( z=0 \) case!

Question 10. Find the square roots of
(i) \( 4 + 3i \)
(ii) \( -6 + 8i \)
(iii) \( -5 – 12i \)
Answer:
(i) To find the square root of \( 4+3i \), we assume the square root is of the form \( a+ib \), where \( a \) and \( b \) are real numbers.
Let \( \sqrt{4+3i} = a+ib \)
Squaring both sides:
\( (a+ib)^2 = 4+3i \)
\( a^2 + 2abi + (ib)^2 = 4+3i \)
\( a^2 - b^2 + 2abi = 4+3i \)
Equating the real and imaginary parts:
\( a^2 - b^2 = 4 \) .......... (1)
\( 2ab = 3 \) .......... (2)
We also use the property \( |z|^2 = |a+ib|^2 = a^2+b^2 \).
\( |4+3i| = \sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5 \)
So, \( a^2+b^2 = 5 \) .......... (3)
Now, we solve the system of equations (1) and (3):
Add (1) and (3):
\( (a^2 - b^2) + (a^2 + b^2) = 4 + 5 \)
\( 2a^2 = 9 \implies a^2 = \frac{9}{2} \implies a = \pm \frac{3}{\sqrt{2}} \)
Subtract (1) from (3):
\( (a^2 + b^2) - (a^2 - b^2) = 5 - 4 \)
\( 2b^2 = 1 \implies b^2 = \frac{1}{2} \implies b = \pm \frac{1}{\sqrt{2}} \)
From equation (2), \( 2ab = 3 \). Since 3 is positive, \( a \) and \( b \) must have the same sign (both positive or both negative).
Therefore, the square roots are \( \pm \left( \frac{3}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) \). This systematic approach helps in finding complex square roots.
(ii) To find the square root of \( -6 + 8i \), we assume the square root is \( a+ib \).
Let \( \sqrt{-6+8i} = a+ib \)
Squaring both sides:
\( (a+ib)^2 = -6+8i \)
\( a^2 - b^2 + 2abi = -6+8i \)
Equating real and imaginary parts:
\( a^2 - b^2 = -6 \) .......... (1)
\( 2ab = 8 \) .......... (2)
Calculate the modulus of \( -6+8i \):
\( |-6+8i| = \sqrt{(-6)^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10 \)
So, \( a^2+b^2 = 10 \) .......... (3)
Now, we solve the system of equations (1) and (3):
Add (1) and (3):
\( (a^2 - b^2) + (a^2 + b^2) = -6 + 10 \)
\( 2a^2 = 4 \implies a^2 = 2 \implies a = \pm \sqrt{2} \)
Subtract (1) from (3):
\( (a^2 + b^2) - (a^2 - b^2) = 10 - (-6) \)
\( 2b^2 = 16 \implies b^2 = 8 \implies b = \pm \sqrt{8} = \pm 2\sqrt{2} \)
From equation (2), \( 2ab = 8 \). Since 8 is positive, \( a \) and \( b \) must have the same sign.
Therefore, the square roots are \( \pm (\sqrt{2} + i 2\sqrt{2}) \). This method ensures all conditions are met.
(iii) To find the square root of \( -5 - 12i \), we assume the square root is \( a+ib \).
Let \( \sqrt{-5-12i} = a+ib \)
Squaring both sides:
\( (a+ib)^2 = -5-12i \)
\( a^2 - b^2 + 2abi = -5-12i \)
Equating real and imaginary parts:
\( a^2 - b^2 = -5 \) .......... (1)
\( 2ab = -12 \) .......... (2)
Calculate the modulus of \( -5-12i \):
\( |-5-12i| = \sqrt{(-5)^2 + (-12)^2} = \sqrt{25+144} = \sqrt{169} = 13 \)
So, \( a^2+b^2 = 13 \) .......... (3)
Now, we solve the system of equations (1) and (3):
Add (1) and (3):
\( (a^2 - b^2) + (a^2 + b^2) = -5 + 13 \)
\( 2a^2 = 8 \implies a^2 = 4 \implies a = \pm 2 \)
Subtract (1) from (3):
\( (a^2 + b^2) - (a^2 - b^2) = 13 - (-5) \)
\( 2b^2 = 18 \implies b^2 = 9 \implies b = \pm 3 \)
From equation (2), \( 2ab = -12 \). Since -12 is negative, \( a \) and \( b \) must have opposite signs.
Therefore, the square roots are \( \pm (2 - 3i) \). This consistent method helps to accurately find complex square roots.In simple words: For each complex number, we set its square root equal to \( a+ib \). Then we square both sides and separate the real and imaginary parts to form two equations. We also use the modulus of the original complex number to get a third equation. By solving these equations, we find the values of \( a \) and \( b \). Remember to check the sign of \( 2ab \) to make sure \( a \) and \( b \) have the correct signs (same sign if positive, opposite signs if negative).

🎯 Exam Tip: Always make sure the signs of \( a \) and \( b \) in your final answer match the sign of \( 2ab \) from your initial equations. If \( 2ab \) is positive, \( a \) and \( b \) have the same sign; if negative, they have opposite signs.