Class 11 Maths NCERT Solutions Chapter 4 Miscellaneous Exercise

i^{18}=i^{4(4)+2}=i^2=-1

i^{25}=i^{4(6)+1}=i,\;\text{so}\;\left(\dfrac{1}{i}\right)^{25}=\dfrac{1}{i^{25}}=\dfrac{1}{i}=-i

i^{18}+\left(\dfrac{1}{i}\right)^{25}=-1+(-i)=-1-i

Now cube (−1 − i) = −(1 + i):

(1+i)^2=1+2i+i^2=2i

(1+i)^3=(1+i)(2i)=2i+2i^2=-2+2i

So (-1-i)^3=-(1+i)^3=-(-2+2i)=2-2i

Let z₁ = a + ib and z₂ = c + id, so Re z₁ = a, Im z₁ = b, Re z₂ = c, Im z₂ = d.

z_1z_2=(a+ib)(c+id)=(ac-bd)+i(ad+bc)

So Re(z₁z₂) = ac − bd.

\dfrac{1}{1-4i}=\dfrac{1+4i}{1^2+4^2}=\dfrac{1+4i}{17}

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

Subtracting:

\dfrac{1+4i}{17}-(1-i)=\dfrac{1+4i-17(1-i)}{17}=\dfrac{1+4i-17+17i}{17}=\dfrac{-16+21i}{17}

\dfrac{3-4i}{5+i}=\dfrac{(3-4i)(5-i)}{5^2+1^2}=\dfrac{15-3i-20i+4i^2}{26}=\dfrac{11-23i}{26}

\dfrac{-16+21i}{17}\times\dfrac{11-23i}{26}=\dfrac{(-16+21i)(11-23i)}{442}

Expanding the numerator:

(-16)(11)+(-16)(-23i)+(21i)(11)+(21i)(-23i)

=-176+368i+231i-483i^2=-176+599i+483=307+599i

Squaring both sides of the given relation:

(x-iy)^2=\dfrac{a-ib}{c-id}

Taking the modulus of both sides, and using |z₁/z₂| = |z₁|/|z₂| and |zⁿ| = |z|ⁿ:

|x-iy|^2=\dfrac{|a-ib|}{|c-id|}

x^2+y^2=\dfrac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}

Squaring both sides:

z_1+z_2+1=(2-i)+(1+i)+1=4

z_1-z_2+1=(2-i)-(1+i)+1=2-2i

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

|1+i|=\sqrt{1^2+1^2}=\sqrt2

Taking the modulus of both sides (2x² + 1 is a positive real number, so it comes out unchanged):

|a+ib|=\dfrac{|x+i|^2}{2x^2+1}

|x+i|^2=x^2+1

So:

|a+ib|=\dfrac{x^2+1}{2x^2+1}

Squaring both sides, and using |a+ib|² = a² + b²:

z̄₁ = 2 + i

z_1z_2=(2-i)(-2+i)=-4+2i+2i-i^2=-4+4i+1=-3+4i

\dfrac{-3+4i}{2+i}=\dfrac{(-3+4i)(2-i)}{2^2+1^2}=\dfrac{-6+3i+8i-4i^2}{5}=\dfrac{-6+11i+4}{5}=\dfrac{-2+11i}{5}

z_1\bar z_1=|z_1|^2=2^2+(-1)^2=5

So 1/(z₁z̄₁) = 1/5, which is a purely real number.

The conjugate of −6 − 24i is −6 + 24i.

Expand the given product:

(x-iy)(3+5i)=3x+5xi-3iy-5i^2y=(3x+5y)+i(5x-3y)

Setting this equal to −6 + 24i and comparing real and imaginary parts:

3x+5y=-6 \qquad 5x-3y=24

Multiply the first equation by 3 and the second by 5, then add to eliminate y:

9x+15y=-18

25x-15y=120

Adding: 34x = 102, so x = 3.

Substituting back: 3(3) + 5y = −6 ⟹ 5y = −15 ⟹ y = −3

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

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

i-(-i)=2i

Expand (x + iy)³:

(x+iy)^3=x^3+3x^2(iy)+3x(iy)^2+(iy)^3=x^3+3ix^2y-3xy^2-iy^3

=(x^3-3xy^2)+i(3x^2y-y^3)

So u = x³ − 3xy² and v = 3x²y − y³.

\dfrac{u}{x}=\dfrac{x^3-3xy^2}{x}=x^2-3y^2

\dfrac{v}{y}=\dfrac{3x^2y-y^3}{y}=3x^2-y^2

Adding:

\dfrac{u}{x}+\dfrac{v}{y}=(x^2-3y^2)+(3x^2-y^2)=4x^2-4y^2

Since |β| = 1, we have \beta\bar\beta=1.

Compute |β − α|²:

|\beta-\alpha|^2=(\beta-\alpha)(\bar\beta-\bar\alpha)=\beta\bar\beta-\beta\bar\alpha-\alpha\bar\beta+\alpha\bar\alpha=1-\beta\bar\alpha-\alpha\bar\beta+|\alpha|^2

Compute |1 − ᾱβ|²:

|1-\bar\alpha\beta|^2=(1-\bar\alpha\beta)(1-\alpha\bar\beta)=1-\alpha\bar\beta-\bar\alpha\beta+\bar\alpha\beta\alpha\bar\beta

Since |β|² = ββ̄ = 1, the last term simplifies: \bar\alpha\beta\alpha\bar\beta=\alpha\bar\alpha\cdot\beta\bar\beta=|\alpha|^2

So |1-\bar\alpha\beta|^2=1-\alpha\bar\beta-\bar\alpha\beta+|\alpha|^2

Comparing the two expressions — they are identical, since β̄α = αβ̄ under reordering of multiplication is not needed here; both expressions contain the same four terms.

|1-i|=\sqrt{1^2+(-1)^2}=\sqrt2

So the equation becomes:

(\sqrt2)^x=2^x \;\Rightarrow\; 2^{x/2}=2^x

Equating exponents:

\dfrac{x}{2}=x \;\Rightarrow\; x=0

The only solution is x = 0, which is excluded since the question asks for non-zero integral solutions.

Taking the modulus of both sides, and using |z₁z₂z₃z₄| = |z₁||z₂||z₃||z₄|:

|a+ib|\,|c+id|\,|e+if|\,|g+ih|=|A+iB|

Squaring both sides:

|a+ib|^2|c+id|^2|e+if|^2|g+ih|^2=|A+iB|^2

Simplify (1+i)/(1−i) first:

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

So the equation becomes:

i^m=1

This holds exactly when m is a multiple of 4, since i⁴ = 1 is the smallest such power.