Question

pls guys can u give me 18 19 and 20 answer for reference 17 answer is also there in the pic pls guys give me the answer ​

Answer 1

$\pink{z = {( \frac{1}{3} + 3i})^{3} } \\ \\ z = { (\frac{1}{3} )}^{3} + 3 { (\frac{1}{3}) }^{2} 3i + 3 \frac{1}{3} {(3i)}^{2} + {(3i)}^{3} \\ \\ z = \frac{1}{27} +3 i + 9 {i}^{2} + 27 {i}^{3} \\ \\ z = \frac{1}{27} + 3i - 9 - 27i \\ \\ z = \frac{1}{27} - 9 - 24i \\ \\ z = \frac{1 - 243}{27} - 24i \\ \\ z = \frac{ - 242}{22} - 24i$

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Answer 2

Solution of question 18 :-

$\implies \sf z = { ( \dfrac{1}{3} + 3i \large)}^{3}$

Where :- i = √-1

We know that :-

$\underline{\boxed{{ \large\bf ( a + b)}^{3} = \large\bf{a}^{3} + {b}^{3} + 3 {a}^{2} b + 3a {b}^{2} }}$

Therefore :-

$\implies \sf { ( \dfrac{1}{3} + 3i \large)}^{3} = { (\dfrac{1}{3}) }^{3} + {(3i)}^{3} + 3 {( \dfrac{1}{3}) }^{2} 3i + 3 {(3i) }^{2} \dfrac{1}{3}$

$\implies \sf \dfrac{1}{27} + 27{(i)}^{3} + i + 9 {(i) }^{2}$

Now we know that :-

• i² = -1
• i³= -i

$\implies \sf \dfrac{1}{27} - 27i + i - 9$

$\implies \sf \dfrac{1}{27} - 9- 26i$

$\implies \sf \dfrac{1 - 243}{27} - 26i$

$\implies \sf \dfrac{ - 242}{27} - 26i$

Solution of question 19 :-

$\implies \sf z = { ( - 2 - \dfrac{1}{3}i )}^{3}$

$\implies \sf z = { (- 1)}^{3} { ( 2 + \dfrac{1}{3}i )}^{3}$

$\implies \sf z = - { ( 2 + \dfrac{1}{3}i )}^{3}$

We know that :-

$\underline{\boxed{{ \large\bf ( a + b)}^{3} = \large\bf{a}^{3} + {b}^{3} + 3 {a}^{2} b + 3a {b}^{2} }}$

$\implies \sf - ( {2 + \dfrac{1}{3}i )}^{3} = - ( \: \: \: { 2 }^{3} + {( \dfrac{1}{3}i)}^{3} + 3 {( 2) }^{2} \dfrac{1}{3} i + 3 {(\dfrac{1}{3}i) }^{2} 2 \: \: \: )$

$\implies \sf - ( 8 + \dfrac{1}{27} {(i)}^{3} + 4 i + {\dfrac{2}{3} {i}^{2} } )$

Now we know that :-

• i² = -1
• i³= -i

$\implies \sf - ( 8 - \dfrac{1}{27} i+ 4 i - {\dfrac{2}{3} } )$

$\implies \sf - 8 + \dfrac{1}{27} i - 4 i + {\dfrac{2}{3} }$

$\implies \sf {\dfrac{2}{3} } - 8 + \dfrac{1}{27} i - 4 i$

$\implies \sf {\dfrac{2 - 24}{3} } + \dfrac{i - 108i}{27}$

$\implies \sf {\dfrac {- 22}{3} } - \dfrac{107i}{27}$

Solution of question 20 :-

$\implies \sf \dfrac{(3 + i\sqrt{5})(3 - i\sqrt{5}) }{( \sqrt{3} + i\sqrt{2}) -( \sqrt{3} - i\sqrt{2}) }$

We know that :-

$\underline{\boxed{{ \large\bf( a+ b)( a- b ) = {a}^{2} - {b}^{2} }}}$

therefore :-

$\implies \sf \dfrac{( {3)}^{2} - {(i\sqrt{5})}^{2} }{\sqrt{3} + i\sqrt{2}-\sqrt{3} + i\sqrt{2} }$

$\implies \sf \dfrac{9 - 5 {(i)}^{2} }{ 2\sqrt{2} i }$

Now we know that :-

• i² = -1

$\implies \sf \dfrac{9 + 5 }{ 2\sqrt{2} i }$

$\implies \sf \dfrac{14 }{ 2\sqrt{2} i }$

$\implies \sf \dfrac{7 }{ \sqrt{2} \: i }$

$\implies \sf \dfrac{7 }{ \sqrt{2} \: i } \times \dfrac{ \sqrt{2} \: i }{ \sqrt{2} \: i }$

$\implies \sf - \dfrac{7 \sqrt{2} \: i }{ {2} }$