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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5

Question 1.

Find the modulus of the following complex numbers.

(i) \(\frac{2 i}{3+4 i}\)

(ii) \(\frac{2-i}{1+i}+\frac{1-2 i}{1-i}\)

(iii) (1 – i)^{10}

(iv) 2i(3 – 4i) (4 – 3i)

Solution:

(iii) (1 – i)^{10}

Solution:

|z| = |\(\sqrt{1^2+1^2}\)|^{10} [∵ |z^{n}| = |z|^{n}]

= (√2)^{10} = 2^{5} = 32

(iv) 2i(3 – 4i)(4 – 3i)

Solution:

= |2i||3 – 4i||4 – 3i|

= 2i(12 – 9i – 16i – 12)

= 2i(-25i) = 50

∴ |z| = 50

Question 2.

For any two complex numbers z_{1} and z_{2}, such that |z_{1}| = |z_{2}| = 1 and z_{1} z_{2} ≠ -1, then show that \(\frac{z_{1}+z_{2}}{1+z_{1} z_{2}}\) is a real number.

Solution:

|z_{1}|^{2} = 1

⇒ \(z_{1} \bar{z}_{1}=1\)

⇒ \(z_{1}=\frac{1}{\bar{z}_{1}}\)

Similarly \(z_{2}=\frac{1}{\bar{z}_{2}}\)

Question 3.

Which one of the points 10 – 8i, 11 + 6i is closest to 1 + i.

Solution:

Let z_{1 }= 1 + i

z_{2 }= 10 – 8i

z_{3} = 11 + 6i

|z_{1}– z_{2}| = |1 + i – 10 + 8i|

= |-9 + 9i|

= \(\sqrt{(-9^2)+(9^2)}\)

= \(\sqrt{162}\) = 9√2

|z_{1}– z_{3}| = |1 + i – 11 – 6i|

= |-10 – 5i|= \(\sqrt{(-10^2) + (-5^2)}\)

= \(\sqrt{100+25}\)

= \(\sqrt{125}\)

= \(\sqrt{5×25}\) = 5√5

∴ 5√5 < 9√2

∴ |z_{1}– z_{2}| > |z_{1}– z_{3}|

∴ 11 + 6i is closest to 1 + i

Question 4.

If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13.

Solution:

|z| = 3, To find the lower bound and upper bound we have

||z_{1}| – |z_{2}|| ≤ |z_{1} + z_{2}| ≤ |z_{1}| + |z_{2}|

||z| – |6 – 8i|| ≤ |z + 6 – 8i| ≤ |z| + |6 – 8i|

|3 – \(\sqrt{36+64}\)| ≤ |z + 6 – 8i| ≤ 3 + \(\sqrt{36+64}\)

|3 – 10| ≤ |z + 6 – 8i| ≤ 3 + 10

7 ≤ |z + 6 – 8i| ≤ 13

Question 5.

If |z| = 1, show that 2 ≤ |z^{2} – 3| ≤ 4.

Solution:

given |z| = 1

Now |z² – 3| ≤ |z²| + |-3|

= |z|² + 3

= 1 + 3 = 4

∴ |z² – 3| ≤ 4 ……….. (1)

|z² – 3| ≥ ||z²|-|-3||

= |1 – 3| = |-2| = 2

|z² – 3| ≥ 2 ………. (2)

From 1 and 2

2 ≤ |z² – 3| ≤ 4

Hence Proved.

Question 6.

If \(\left|z-\frac{2}{z}\right|\) = 2, show that the greatest and least value of |z| are √3 + 1 and √3 – 1 respectively.

Solution:

\(\left|z-\frac{2}{z}\right|\) = 2

We know that

The minimum value of |z| is |1 – √3| = √3 – 1

The greatest value of |z| is √3 + 1

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.

Solution:

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|.

Solution:

The vertices are z = x + iy.

iz = i(x + iy) = -y + ix

z + iz = (x – y) + i(x + y)

the points representing the complex numbers z, iz and z + iz are (x, y), (-y, x) and ((x – y), (x + y)) respectively.

given area of triangle = 50 square units

|\(\frac {1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]| = 50

|x(x – x – y) -y(x + y – y) + (x – y) (y – x)| = 100

|-xy – xy + 2xy – x² – y²| = 100

x² + y² = 100

|z|² = 100

|z| = 10

Question 9.

Show that the equation z^{3} + 2\(\bar{z}\) = 0 has five solutions.

Solution:

Given that z^{3} + 2\(\bar{z}\) = 0

z^{3} = -2 \(\bar{z}\) ……. (1)

Taking modulus on both sides,

z has four non-zero solution.

Hence including zero solution. There are five solutions.

Question 10.

Find the square roots of

(i) 4 + 3i

(ii) -6 + 8i

(iii) -5 – 12i

Solution:

(i) z = 4 + 3i

|z| = |4 + 3i| = \(\sqrt{16+9}\) = 5

### Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.5 Additional Problems

Question 1.

Find the modulus and argument of the following complex numbers and convert them in polar form.

Solution:

(i)

Question 2.

Find the square roots of -15 – 8i

Solution:

On solving (i) and (iii), we get

x^{2} = 1 and y^{2} = 16

⇒ x = ± 1 and y = ±4

From (ii), we observe that 2xy is negative.

So, x and y are of opposite signs.

Question 3.

Find the square roots of i.

Solution:

From (ii) we observe that we find that 2xy is positive. So, x and y are of same sign.

Question 4.

Find the modulus or the absolute value of

Solution:

Question 5.

Find the modulus and argument of the following complex numbers:

Solution:

Question 6.

Show that the points representing the complex numbers 7 + 9i, – 3 + 7i, 3 + 3i form a right-angled triangle on the Argand diagram.

Solution:

Let A, B and C represent the complex numbers

7 + 9i, – 3 + 7i and 3 + 3i in the Argand diagram respectively.

Hence ∆ABC is a right-angled isosceles triangle.

Question 7.

Find the square root of (-7 + 24i).

Solution: