Question 7, Exercise 1.1

Solutions of Question 7 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the magnitude of the $11+12 i$.

Solution.

Suppose $$z=11+12i$$ Then $$\begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265}\end{align}$$ Hence $|11+12 i|=\sqrt{265}$.

Find the magnitude of the $(2+3 i)-(2+6 i)$.

Solution. Suppose $z=(2+3i)−(2+6i)$, then $$\begin{align}z&=2+3i−2−6i\\ &=-3i \end{align}$$ Now $$\begin{align} |z| &= \sqrt{0^2+(-3)^2} \\ &= \sqrt{9} = 3. \end{align}$$ Hence $$|(2+3 i)-(2+6 i)|=3.$$

Find the magnitude of the $(2-i)(6+3 i)$.

Solution. Suppose $$z=(2-i)(6+3 i),$$ then $$\begin{align} |z|&=|(2-i)(6+3 i)|\\ &=|(2-i)| |(6+3 i)|\\ &=\sqrt{2^2+1^2} \sqrt{6^2+3^2} \\ &=\sqrt{5}\sqrt{45}\\ &=\sqrt{5}\cdot 3\sqrt{5} = 15.\end{align}$$ Hence $|(2-i)(6+3 i)|=15$.

Find the magnitude of the $\dfrac{3-2 i}{2+i}$.

Solution.

Suppose $$z=\dfrac{3-2 i}{2+i},$$ then $$\begin{align} |z|&=\left|\dfrac{3-2 i}{2+i} \right| \\ &=\dfrac{|3-2 i|}{|2+i|}\\ &=\dfrac{\sqrt{3^2+(-2)^2}}{\sqrt{2^2+1^2}}\\ &=\dfrac{\sqrt{13}}{\sqrt{5}} = \sqrt{\dfrac{13}{5}}.\end{align}$$ Hence $\left|\dfrac{3-2 i}{2+i} \right|=\sqrt{\dfrac{13}{5}}$.

Find the magnitude of the $(\sqrt{3}-\sqrt{-8})(\sqrt{3}+\sqrt{-8})$.

Solution. Suppose $$\begin{align}z&=(\sqrt{3}-\sqrt{-8})(\sqrt{3}+\sqrt{-8}) \\ &=(\sqrt{3}-\sqrt{8}i)(\sqrt{3}+\sqrt{8}i) \\ &=(\sqrt{3})^2-(\sqrt{8}i)^2\\ &=3+8 = 11.\end{align}$$ Then $$|z|=|11|=11.$$ Hence $|(\sqrt{3}-\sqrt{-8})(\sqrt{3}+\sqrt{-8})|=11$.

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