Class 11 Maths NCERT Solutions Chapter 4 Ex 4.1 – Algebra of Complex Numbers

(5i)\left(-\dfrac{3}{5}i\right)=5\times\left(-\dfrac{3}{5}\right)\times i^2=-3\times(-1)=3

i^9=i^{4(2)+1}=i

i^{19}=i^{4(4)+3}=i^3=-i

i^9+i^{19}=i+(-i)=0

−39 = 4(−10) + 1, so i^{-39}=i^{4(-10)+1}=i

3(7+i7)+i(7+i7)=21+21i+7i+7i^2

=21+21i+7i-7=(21-7)+(21+7)i

(1-i)-(-1+i6)=1-i+1-6i=2-7i

Real part: \dfrac{1}{5}-4=\dfrac{1}{5}-\dfrac{20}{5}=-\dfrac{19}{5}

Imaginary part: \dfrac{2}{5}-\dfrac{5}{2}=\dfrac{4}{10}-\dfrac{25}{10}=-\dfrac{21}{10}

First add the inner bracket:

\left(\dfrac{1}{3}+4\right)+i\left(\dfrac{7}{3}+\dfrac{1}{3}\right)=\dfrac{13}{3}+i\dfrac{8}{3}

Now subtract (−4/3 + i):

\left(\dfrac{13}{3}+\dfrac{4}{3}\right)+i\left(\dfrac{8}{3}-1\right)=\dfrac{17}{3}+i\dfrac{5}{3}

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

(1-i)^4=(-2i)^2=4i^2=-4

Using (a+b)^3=a^3+3a^2b+3ab^2+b^3 with a = 1/3, b = 3i:

a^3=\dfrac{1}{27}

3a^2b=3\times\dfrac{1}{9}\times3i=i

3ab^2=3\times\dfrac{1}{3}\times9i^2=1\times(-9)=-9

b^3=27i^3=27(-i)=-27i

Adding: real part = \dfrac{1}{27}-9=\dfrac{1-243}{27}=-\dfrac{242}{27}; imaginary part = 1-27=-26

Let z = −2 − i/3. First square z:

z^2=(-2)^2+2(-2)\left(-\dfrac{i}{3}\right)+\left(-\dfrac{i}{3}\right)^2=4+\dfrac{4i}{3}-\dfrac{1}{9}=\dfrac{35}{9}+\dfrac{4i}{3}

Now multiply by z again:

z^3=\left(\dfrac{35}{9}+\dfrac{4i}{3}\right)\left(-2-\dfrac{i}{3}\right)

=-\dfrac{70}{9}-\dfrac{35i}{27}-\dfrac{8i}{3}-\dfrac{4i^2}{9}

=-\dfrac{70}{9}+\dfrac{4}{9}+i\left(-\dfrac{35}{27}-\dfrac{72}{27}\right)=-\dfrac{66}{9}-\dfrac{107}{27}i

z = 4 − 3i, so z̄ = 4 + 3i and |z|² = 4² + (−3)² = 16 + 9 = 25

z^{-1}=\dfrac{\bar z}{|z|^2}=\dfrac{4+3i}{25}

z = √5 + 3i, so z̄ = √5 − 3i and |z|² = (√5)² + 3² = 5 + 9 = 14

z^{-1}=\dfrac{\bar z}{|z|^2}=\dfrac{\sqrt5-3i}{14}

z = −i = 0 − i(1), so z̄ = i and |z|² = 0² + (−1)² = 1

z^{-1}=\dfrac{\bar z}{|z|^2}=\dfrac{i}{1}=i

Check: (−i)(i) = −i² = 1 ✓

(3+i\sqrt5)(3-i\sqrt5)=3^2-(i\sqrt5)^2=9-(-5)=14

(\sqrt3+\sqrt2i)-(\sqrt3-\sqrt2i)=2\sqrt2i

\dfrac{14}{2\sqrt2i}=\dfrac{7}{\sqrt2i}

Using 1/i = −i:

\dfrac{7}{\sqrt2}\times\dfrac{1}{i}=\dfrac{7}{\sqrt2}\times(-i)=-\dfrac{7i}{\sqrt2}

Rationalizing the denominator:

-\dfrac{7i}{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=-\dfrac{7\sqrt2}{2}i