1 Marks Question

Question 11 Mark

For a positive integer n, find the value of $(1-\text{i})^{\text{n}}\Big(1-\frac{1}{\text{i}}\Big)^{\text{n}}$

Answer

We have$(1-\text{i})^{\text{n}}\Big(1-\frac{1}{\text{i}}\Big)^{\text{n}}$
$=\Big[(1-\text{i})^{\text{n}}\Big(1-\frac{1}{\text{i}}\Big)\Big]^{\text{n}}$
$=\Big[(1-\text{i})\Big(1-\frac{1}{\text{i}}\times\frac{\text{i}}{\text{i}}\Big)\Big]^{\text{n}}$
$=\Big[(1-\text{i})\Big(1-\frac{\text{i}}{\text{i}^2}\Big)\Big]^{\text{n}}$ $[\because\text{ i}^2=-1\big]$
$=\big[(1-\text{i})(1+\text{i})\big]^{\text{n}}$
$=\big[1-\text{i}^2\big]^{\text{n}}$
$=\big[1+1\big]^{\text{n}}$
$=2^{\text{n}}$
Hence,$(1-\text{i})^{\text{n}}\Big(1-\frac{1}{\text{i}}\Big)^{\text{n}}=2^{\text{n}}$

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Question 21 Mark

Find $\Big|(1+\text{i})\frac{(2+\text{i})}{(3+\text{i})}\Big|$

Answer

$\Big|(1+\text{i})\frac{(2+\text{i})}{(3+\text{i})}\Big|=|1+\text{i}|\frac{|2+\text{i}|}{|3+\text{i}|}$
$=\sqrt{1^2+1^2}\frac{\sqrt{2^2+1^2}}{\sqrt{3^2+1^2}}$
$=\sqrt{2}\frac{\sqrt{5}}{\sqrt{10}}=1$

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Question 31 Mark

If $(1+\text{i})\text{z}=(1-\text{i})\bar{\text{z}},$ then show that $\text{z}=-\text{i}\bar{\text{z}}$

Answer

We have, $(1+\text{i})\text{z}=(1-\text{i})\bar{\text{z}}$
$\text{z}=\frac{1-\text{i}}{1+\text{i}}\bar{\text{z}}=\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\bar{\text{z}}$
$=\frac{(1-\text{i})^2}{(1-\text{i}^2)}\bar{\text{z}}=\frac{1-2\text{i}+\text{i}^2}{1+1}\bar{\text{z}}$
$=\frac{1-2\text{i}-1}{2}\bar{\text{z}}=-\text{i}\bar{\text{z}}$

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