Modulus of Complex Number MCQ Quiz - Objective Question with Answer for Modulus of Complex Number - Download Free PDF

Explanation:

\(\rm \frac{1}{2}+Re\left(\frac{Z_1}{Z_2}\right) = \frac{1}{2}+ Re (\frac{\omega }{\omega^2})\)

= \(\frac{1}{2}Re(\omega)^2\)

= \(\frac{1}{2}+ Re [ -\frac{1}{2} - \frac{\sqrt3}{2}i\)

= \(\frac{1}{2} + (-\frac{1}{2}) = 0\)

∴ Option (b) is correct.

Explanation:

\(\rm Z_1^2+Z_2^2+Z_1Z_2=0\)

⇒ Z ₁  = ω and Z ₂  = ω ²

Now

⇒ \(|\frac{Z_1}{Z_2}| =| \frac{\omega }{(\omega)^2}| =|\frac{1}{\omega|}= 1\)

Option (a) is correct.

Explanation:

Given

 \(\rm z=\frac{1}{3}\begin{vmatrix}i&2i&1\\\ 2i&3i&2\\\ 3&1&3\end{vmatrix}\) 

\(Z = \frac{1}{3}[[ (i9 – 2) – 2i (6i – 6) 1(2i – 9i)]\)

\(Z = \frac{1}{3}[3+3i] =1 + i \)

|Z| = \(√{1^2+1^2} =\) √2 

∴ Option (b) is correct.

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

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

Given:

|z| = 4 and arg z = \(\displaystyle \frac{5π}{6}\)

Let z = |z| (cos θ + i sin θ) where θ = arg(z)

⇒ z = \(\displaystyle 4\left(cos\frac{5\pi}{6}+i \ sin\frac{5\pi}{6}\right)\)

⇒ z = \(\displaystyle 4\left(\frac{-√3}{2}+i \ \frac{1}{2}\right)\)

⇒ z = - 2√3 + 2i

∴ z = - 2√3 + 2i

Concept:

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

Calculations:

Let \(\rm z= x + iy = \dfrac{4+2i}{1-2i}\)

\(\rm = \dfrac{4+2i}{1-2i}\times\dfrac{1+2i}{1+2i}\)

\(\rm= \dfrac{4+10i+4i^2}{1-4i^2}\)   

As we know i2 = -1 

\(\rm = \dfrac{4+10i-4}{1+4}\)

\(\rm x + iy =\dfrac{10i}{5} = 0 + 2i\)

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| = \(\rm \sqrt{0^2+2^2} = 2\)

Concept;

Modulus of a complex number z = x + iy is given by:

|z| = \(\rm \sqrt{x^2 + y^2}\)

(a + b)(a - b) = a² - b²

Calculation: 

If z = \(\rm \frac{z_1}{z_2}\) so, modulus of |z| = \(\rm \frac{|z_1|}{|z_2|}\)

z = \(\rm \frac {1+i}{1+\sqrt3 i}\)

|z| = \(\rm \frac {\sqrt {(1)^2 + (1)^2}}{\sqrt {(1)^2 + (\sqrt 3)^2}} = \frac{\sqrt 2}{2} = \frac {1}{\sqrt 2}\) 

Concept:

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

Calculations:

Let \(\rm z = x+iy =\left(\frac{1+i}{1-i} - \frac{1-i}{1+i}\right)\)

\(\rm =\frac{(1+i)^2-(1-i)^2}{1^2-i^2}\\=\frac{1+2i+i^2-1+2i-i^2}{1+1}\\=\frac{4i}{2}=2i\)

z = x + iy = 0 + 2i

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| = \(\rm \sqrt{0^2+2^2} = 2\)

Concept:

Iota power:

• i² = -1, i³ = -i, i⁴ = 1, i⁴ⁿ = 1
• Number of the form 2n is always even, n ∈ N
• Numebr of the form 2n +1 or 2n - 1 is always odd, n ∈ N
• (-a)^{2n -1} = -(a)^{2n -1}

Calculation:

Let,

Z = i2n + 1(-i)2n - 1

⇒ Z = i2n+1 × (i)2n-1 × (-1)^2n-1

⇒ Z =  i2n +1+2n-1 ×(-1)^2n-1

⇒ Z = i⁴n × (-1)          [∵ (-1)2n-1 = -1]

⇒ Z = (-1)⁴ⁿ ×

⇒ Z = i4n × (-1)

⇒ Z = -(i)4n × (-1)

⇒  Z = 1 × (-1)         [∵ i⁴ⁿ = 1] 

⇒ Z = -1

⇒ |Z| = |-1| = 1

Hence, the modulus of the complex number i2n + 1(-i)^{2n - 1} is 1.

Concept;

Modulus of a complex number z = x + iy is given by:

|z| = \(\rm \sqrt{x^2 + y^2}\)

(a + b)(a - b) = a² - b²

Calculation:

Given: z = \(\rm \frac {5- 5i}{3 - 4i}\)

Mutiply by (3 + 4i) in denominator and numerator.

z = \(\rm (\frac {5- 5i}{3 - 4i}) \times (\frac{3 + 4i}{3 + 4i})\) = \(\rm 5[\frac {(1- i)(3+4i)}{(3)^2 - (4i)^2}]\) = \(\rm 5(\frac{3 + 4i - 3i - 4i^2}{25})\)                  (∵ i² = -1) 

z = \(\rm \frac {7 + i}{5}\)

|z| = \(\rm \frac{\sqrt{(7)^2 + (1)^2}}{5}= \frac{\sqrt{49 +1}}{5} = \frac{√{50}}{5} = \frac{5√2}{5}\)

|z| = √2 

If z = \(\rm \frac{z_1}{z_2}\) so, modulus of |z| = \(\rm \frac{|z_1|}{|z_2|}\)

z = \(\rm \frac {5- 5i}{3 - 4i}\)

|z| = \(\rm \frac {\sqrt {(5)^2 + (5)^2}}{\sqrt {(3)^2 + (4)^2}} = \frac{√50}{√25} = \frac {5√2}{5}\) 

|z| = √2  

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:

• i2 = - 1
• If z = x + iy then \(|z| = \sqrt{x^2 + y^2}\) 

CALCULATION:

Given: z = (1 - i)4 First let's simplify the expression (1 - i)⁴ 

⇒ (1 - i)² = 1 + i² - 2i

As we know that, i² = - 1

⇒ (1 + i)² = -2i

Since (1 - i)4 = (1 - i)2 × (1 - i)2 we get:

⇒ (1 + i)⁴ = (-2i)² = - 4

⇒ z = - 4 + 0i

As we know that, if z = x + iy then \(|z| = \sqrt{x^2 + y^2}\) 

Here, x = - 4 and y = 0

⇒ \(|z| = \sqrt{(-4)^2 + 0^2} = \pm 4\) 

As we know that, |z| denotes the distance between origin and z in the argand plane. So, |z| cannot be negative

⇒ |z| = 4

Hence, correct option is 2.

Concept:

The modulus of a complex number z = x + iy

|z| = \(\rm \sqrt{x^2+y^2}\)

The conjugate \(\rm \overline z\) = x - iy

Calculation:

z = 1+ 2i

\(\rm \overline z\)= 1 - 2i

S = \(\rm\left| z + \overline z +1\over z + \overline z -1\right|\)

S = \(\rm\left| 1+2i+1-2i+1\over 1+2i+1-2i -1\right|\)

S = \(\rm\left| 3\over 1\right|\) 

S = 3

Concept:

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

Solution:

The given equation can be solved as:

iz3 + z2 - z + i = 0

⇒ iz3 + i²z + z2 + i = 0

⇒ iz(z² + i) + (z2 + i) = 0

⇒ (z² + i)(zi + 1) = 0

⇒ z² + i = 0 or zi + 1 = 0

⇒ z2 = -i or z = i

|z| = 1 in both the cases.

Concept: 

Modulus of complex no. z =  a + ib is given by |z| = \(\rm \sqrt{a^{2}+ b^{2}}\) . 

Property of complex number:

\(\rm \left|\frac {z_1}{z_2}\right| = \frac {|z_1|}{|z_2|}\)

Calculation:

Let z = \(\rm \frac {\cos \theta - i \sin \theta}{\cos \theta + i \sin \theta}\)

Taking modulus on both sides, we get

⇒ |z| = \(\rm \left|\frac {\cos θ - i \sin θ}{\cos θ + i \sin θ}\right|\)

= \(\rm \frac {|\cos θ - i \sin θ|}{|\cos θ + i \sin θ|}\)

= \(\rm \frac {\sqrt {\cos^2 \theta + \sin^2 \theta}}{\sqrt {\cos^2 \theta + \sin^2 \theta}}\)

= 1