Properties of Complex Numbers MCQ Quiz in मल्याळम - Objective Question with Answer for Properties of Complex Numbers - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

If z = a + ib , |z| = \(\sqrt{a^2\,+\,b^2}\)

If z = a + ib , \(\frac{1}{z}\) = \(\frac{a \,-\,ib}{{a^2\,+\,b^2}}\) 

|z₁z₂| = |z₁| × |z₂|

Calculation:

Given z1 = 1 - 2i , z2 = 1 + i and z3 = 3 + 4i

∴ \(\frac{1}{z_{1}}\) = \(\frac{1\,+\,2i}{{1^2\,+\,2^2}}\) = \(\frac{1\,+\,2i}{{5}}\)

Similarly  \(\frac{2}{z_{2}}\) = 2 ×  \(\frac{1}{z_{2}}\)  = 2 × \(\frac{1\,-\,i}{{1^2\,+\,1^2}}\) = 2 ×  \(\frac{1\,-\,i}{{2}}\) = (1 - i)

\(\frac{2}{z_{2}}\) = (1 - i)

⇒ \(\frac{1}{z_{2}} \) = \(\frac{1-i}{2}\)

\(\frac{z_3}{z_2}\) = z₃ ×  \(\frac{1}{z_{2}} \)  = (3 + 4i) × \(\frac{1-i}{2}\)

⇒ \(\frac{z_3}{z_2}\) =  \(\frac{7+i}{2}\)

 We need to find \(\left| {\left( {\frac{1}{{z_1}} + \frac{2}{{z_2}}}\right)\frac{{z_3}}{{z_2}}} \right| \)

 \(\left| {\left( {\frac{1}{{z_1}} + \frac{2}{{z_2}}}\right)\frac{{z_3}}{{z_2}}} \right| \) = \(\left| {\left( {\frac{1}{{z_1}} + \frac{2}{{z_2}}}\right)} \right| \) × \(\left | \frac{z_3}{z_2} \right|\)

=  \(\left| {\left( {\frac{1\,+\,2i}{{5}} + (1-i)}\right)} \right|\) × \(\Big|\frac{7+i}{2}\Big|\)

= \(\Big|\frac{1+2i+5-5i}{5}\Big|\) × \(\Big|\frac{7+i}{2}\Big|\)

= \(\Big|\frac{6-3i}{5}\Big|\) × \(\Big|\frac{7+i}{2}\Big|\)

= \(\frac{\sqrt{6^2+(-3)^2}}{5} \times \frac{\sqrt{7^2+1^2}}{2}\)

= \(\frac{\sqrt{36+9}}{5} \times \frac{\sqrt{49+1}}{2}\)

= \(\frac{\sqrt{45}}{5} \times \frac{\sqrt{50}}{2} =\frac{\sqrt{45}}{5} \times \frac{5\sqrt2}{2}\)

=  \(\frac{\sqrt{45} \times \sqrt2}{2}\)

= \(\frac{\sqrt{45} }{\sqrt2}\) 

= \(\sqrt \frac{45}{2} \)

Evaluating , we get \(\sqrt \frac{45}{2} \).

Concept:

Let z = x + iy be a complex number, Where x is called real part of complex number or Re (z) and y is called Imaginary part of the complex number or Im (z)

Calculation:

Let two complex numbers z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂

Where x₁ = Re (z₁), x₂ = Re (z₂) and y₁ = Im (z₁), y₂ = Im (z₂).

Now,

z₁z₂ = (x₁ + iy₁) (x₂ + iy₂)

z₁z₂ = (x₁x₂ – y₁y₂)  + i (x₁y₂ + y₁x₂)

∴ Im (z₁z₂) = (x₁y₂ + y₁x₂)

= Re (z₁) Im (z₂) + Im (z₁) Re (z₂)

Concept:

Conjugate of a complex number:

For any complex number z = x + iy  the conjugate z̅ is given by  z̅ = x - iy

Calculation:

z = \(\rm i + 3 \over 2i + 1\)

z = \(\rm { i + 3 \over 2i + 1}\times {-2i+1\over-2i +1}\)

z = \(\rm -2i^2-6i+i+3\over1 - (2i)^2\)

z = \(\rm {5-5i\over1 + 4}\)

z = \(\rm {5(1 - i)\over5}\)

z = 1 - i

Conjugate of z = z̅  = 1 + i

Concept:

Modulus of the Complex Number z = a + ib is given by:

Mod(z) = |z| = \(\sqrt{[Re(z)]^{2}+[Im(z)]^{2}}\)

= \(\sqrt{a^{2}+b^{2}}\)

Solution:

\(\frac{1+2i}{1-(1-i)}\) = \(\frac{1+2i}{i}\) = \(2+\frac{1}{i}\) = \(2+\frac{1}{i}\times \frac{i}{i}\) = \(2+\frac{i}{i^{2}}\) = \(2+\frac{i}{(-1)}\) = 2 - i

Modulus of the complex number z = 2 - i is 

\(\sqrt{2^{2}+(-1)^{2}}\) = \(\sqrt{4+1}\) = \(\sqrt{5}\) 

Concept:

The argument of a complex number z = x + iy

arg(z) = tan^-1\(\rm \left(y\over x\right)\)

The angle is according to the sign of the y and x

• Both positive then angle ∈ [0°, 90°]
• Negative x and positive y then angle ∈ [90°, 180°]

Calculation:

Given complex number z = 2 + i 2√3

arg(z) = tan^-1\(\rm \left(2\sqrt3\over 2\right)\)

arg(z) = tan^-1\(\rm \sqrt3\)

arg(z) = 60° (∵ Both positive then angle ∈ [0°, 90°])

Concept:

Complex Numbers: For a complex number z = a + ib, the following are defined:

• Conjugate: z̅ = a - ib
• arg(z) = θ = \(\rm\tan^{-1}\left(b\over a\right)\).
• arg(z₁.z₂) = arg(z₁) + arg(z₂).

Calculation:

Given that z̅ + iw̅ = 0.

⇒ z̅ = -iw̅

Taking conjugates, we get:

⇒ z = iw

Multiplying by i, we get:

⇒ w = -iz

Now, arg(zw) = π.

⇒ arg(z.(-iz)) = π

⇒ arg(-i.z²) = π

⇒ arg(-i) + 2 arg(z) = π

⇒ \(-\pi\over2\) + 2 arg(z) = π

⇒ 2 arg(z) = \(3\pi\over2\)

⇒ arg(z) = \(3\pi\over4\).

Concept:

Multiplicative Inverse of a complex number, z  = \(\frac{1}{z}\)

Application:

Multiplicative Inverse of z = 1 + i is z^-1 i.e.

Multiplicative Inverse of z = \(\frac{1}{z}\)

Puttng z = 1 + i, we get,

Multiplicative Inverse of 1 + i  = \(\frac{1}{1+i}\)

Rationalizing, 

=  \(\frac{1}{1+i}\times\frac{1-i}{1-i}\)

=  \(\frac{1-i}{1^2-i^2}\)(Since i²=-1)

=  \(\frac{1-i}{2}\)

Concept

The general form of a complex number is z = x + iy. The polar representation of z

is z = r(cos θ + i sin θ). Here, r is the modulus of z and θ is called the amplitude or argument of the complex number. The formula to find the amplitude of a complex

number is: \(\displaystyle \theta =tan^{-1}\frac{y}{x}\) and ∣z∣ = \(\sqrt {x^2+y^2}\)

Calculation

\(\displaystyle \frac{1 + i}{1 - i}\)

⇒ \(\displaystyle \frac{1 + i}{1 - i}\) = \(\displaystyle \frac{1 + i}{1 - i} \times \frac{1 + i}{1 + i}\)

⇒ \(\displaystyle \frac{1 + i}{1 - i}=\frac{(1 + i)^2}{1-i^2}=\frac{(1 + i)^2}{2}=\frac{1 + i^2+2i}{2}=i\)

⇒ \(\displaystyle \frac{1 + i}{1 - i}\) = 0 + i

Now, Argument = \(\displaystyle \theta =tan^{-1}\frac{y}{x}\) = \(\displaystyle tan^{-1}\frac{i}{0}=\frac{π}{2}\)

Modulus is ∣z∣ = \(\sqrt {x^2+y^2}\)  = \(\sqrt{0^2+1^2}=1 \)

∴ Argument and modulus of \(\frac{1 + i}{1 - i}\) are respectively \(\frac{π}{2}\) and 1.

Concept:

Modulus of z = |z| = \(\rm \sqrt {x^2+y^2} = \sqrt {Re (z)^2+Im (z)^2}\)

Calculations:

Let z = x + iy = (1 + i)² = 1² + 2i + i²         (∵ (a + b)² = a² + b² + 2ab)

z = 1 + 2i - 1                                (∵ i² = -1)

∴ z = x + iy = 2i

So, x = 0  and y = 2

As we know that if z = x + iy be any complex number, then its modulus is given by, |z| = \(\rm \sqrt{x^2+y^2}\)

∴ |z| = \(\rm \sqrt{(0)^2+{2}^2} = \sqrt 4 = 2\) 

Concept:

The circle is defined as the locus of a point that moves in a plane such that its distance from a fixed point in that plane is constant.

Equation of circle having a center (h, k) and radius a is  given as,

(x - h)² + (y - k)² = a²

But when a circle passes through the origin, then the equation of the circle is

x² + y² - 2 × h × x - 2 × k × y = 0 ----(1)

a² = h² + k²  -----(2)

Calculation:

Given:

z z̅  + (1 + i) z +(1 - i) z̅  = 0

As we know,

z = x + iy, z̅ = x - iy using these value in the above given equation we get:

(x + iy)(x - iy) + (1 + i)(x + iy) + (1 - i)(x - iy) = 0

x² + y² + x + iy + ix - y + x - iy - ix - y = 0

x² + y² + 2x - 2y = 0 -----(3)

Comparing (3) with (1) we get:

h = -1 and k = 1 which are centre of circle.

By using h, k values in (2) we get the radius of the circle as √2.

The general equation of a non-degenerate conic section is: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A, B and C are all not zero

The above given equation represents a non-degenerate conics whose nature is given below in the table: