Question 3 & 4, Exercise 1.2

Solutions of Question 3 & 4 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

${{z}_{1}}=\sqrt{3}+\sqrt{2}i$, ${{z}_{2}}=\sqrt{2}-\sqrt{3}i$and ${{z}_{3}}=2+3i$, then verify distributive property w.r.t. addition and multiplication.

${{z}_{1}}=\sqrt{3}+\sqrt{2}i$ ${{z}_{2}}=\sqrt{2}-\sqrt{3}i$ ${{z}_{3}}=2+3i$ Distributive property w.r.t. addition and multiplicative. $$\begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&={{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{3}}\\ {{z}_{2}}+{{z}_{3}}&=\sqrt{2}-\sqrt{3}i+2+3i\\ &=\left( \sqrt{2}+2 \right)+\left( 3-\sqrt{3} \right)i\\ L.H.S.&={{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)\\ &=\left( \sqrt{3}+\sqrt{2}i \right)\left( \left( \sqrt{2}+2 \right)+\left( 3-\sqrt{3} \right)i \right)\\ &=\left( \sqrt{3}+\sqrt{2}i \right)\left( \sqrt{2}+2 \right)+\left( \sqrt{3}+\sqrt{2}i \right)\left( 3-\sqrt{3} \right)i\\ &=\left( \sqrt{3}+\sqrt{2}i \right)\left( \sqrt{2}+2 \right)+\left( \sqrt{3}+\sqrt{2}i \right)\left( 3i-\sqrt{3}i \right)\\ &=\left( \sqrt{6}+2\sqrt{3}+2i+2\sqrt{2}i \right)+\left( 3\sqrt{3}i-3\sqrt{2}-3i+\sqrt{6} \right)\\ &=\left( \sqrt{6}+2\sqrt{3}+2i+2\sqrt{2}i \right)+\left( \sqrt{6}-3\sqrt{2}-3i+3\sqrt{3}i \right)\\ &=\left( 2\sqrt{6}-3\sqrt{2}+2\sqrt{3} \right)+\left( 2\sqrt{2}+3\sqrt{3}-1 \right)i\\ {{z}_{1}}{{z}_{2}}&=\left( \sqrt{3}+\sqrt{2}i \right)\left( \sqrt{2}-\sqrt{3}i \right)\\ &=\left( \sqrt{3}+\sqrt{2}i \right)\left( \sqrt{2}-\sqrt{3}i \right)\\ &=\sqrt{6}+\sqrt{6}+2i-3i\\ &=2\sqrt{6}-i\\ {{z}_{1}}{{z}_{3}}&=\left( \sqrt{3}+\sqrt{2}i \right)\left( 2+3i \right)\\ &=2\sqrt{3}-3\sqrt{2}+2\sqrt{2}i+3\sqrt{3}i\\ {{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{3}}&=\left( 2\sqrt{6}-i \right)+\left( 2\sqrt{3}-3\sqrt{2}+2\sqrt{2}i+3\sqrt{3}i \right)\\ &=\left( 2\sqrt{6}+2\sqrt{3}-3\sqrt{2} \right)+\left( 2\sqrt{2}+3\sqrt{3}-1 \right)i\\ L.H.S.&=R.H.S.\end{align}$$

Find the additive and multiplicative inverse of the complex number $5+2i$.

Given $z=5+2i$. Here $a=5$ and $b=2$.

Additive inverse of $z$ is $-z$ and $-z =-5-2i$.

Thus additive inverse of $5+2i$ is $-5-2i$.

Now $$\begin{align}z^{-1} &= \dfrac{a}{{{a}^{2}}+{{b}^{2}}}-\dfrac{b}{{{a}^{2}}+{{b}^{2}}}i\\ &=\dfrac{5}{5^2+2^2}-\dfrac{2}{5^2+2^2}i\\ &=\dfrac{5}{29}-\dfrac{2}{29}i \end{align}$$ Thus multiplicative inverse of $5+2i$ is $\dfrac{5}{29}-\dfrac{2}{29}i$.

Find the additive and multiplicative inverse of the complex number $\left( 7,-9 \right)$.

Given $z=(7,-9)=7-9i$. Here $a=7$ and $b=-9$.

Additive inverse of $z$ is $-z$ and $-z =-7+9i$.

Thus additive inverse of $7-9i$ is $-7+9i$.

Now $$\begin{align}z^{-1} &= \dfrac{a}{{{a}^{2}}+{{b}^{2}}}-\dfrac{b}{{{a}^{2}}+{{b}^{2}}}i\\ &=\dfrac{7}{7^2+(-9)^2}-\dfrac{-9}{7^2+(-9)^2}i\\ &=\dfrac{7}{130}+\dfrac{9}{130}i \end{align}$$ Thus multiplicative inverse of $(7,-9)$ is $\dfrac{7}{130}+\dfrac{9}{130}i$.

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