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This is one of our many quizzes on Imaginary numbers. Click on this to access more such competitive quizzes on Complex Numbers.

Imaginary numbers Quiz 2

mathematics

 Change the complex number $3\left( {4 + i4} \right) + i\left( {4 + i4} \right)$  in the form $a + ib$

 First open the brackets and then perform necessary operations keeping in mind that ${i^2} =  - 1$.

We have: $3\left( {4 + i4} \right) + i\left( {4 + i4} \right)$ $ \\ \Rightarrow 12 + 12i + 4i + 4{i^2} \\ $ Now, ${i^2} =  - 1$ $ \\ 12 + 16i - 4 \\   \Rightarrow 8 + 16i $

 Express $\left[ {\left( {\dfrac{1}{5} + i\dfrac{7}{5}} \right) + \left( {4 + i\dfrac{1}{5}} \right)} \right] - \left( { - \dfrac{4}{5} + i} \right)$ in the form $a + ib$

 Open the brackets and group the similar terms to perform the operations of addition and subtraction. 

We have: $\left[ {\left( {\dfrac{1}{5} + i\dfrac{7}{5}} \right) + \left( {4 + i\dfrac{1}{5}} \right)} \right] - \left( { - \dfrac{4}{5} + i} \right)$ $ \\ \Rightarrow \left[ {\dfrac{1}{5} + i\dfrac{7}{5} + 4 + i\dfrac{1}{5}} \right] + \dfrac{4}{5} - i \\   \Rightarrow \dfrac{{21}}{5} + i\dfrac{8}{5} + \dfrac{4}{5} - i \\   \Rightarrow \dfrac{{25}}{5} + i\dfrac{3}{5} \\   \Rightarrow 5 + i\dfrac{3}{5} \\  $

 Given that $\omega $ is an imaginary cube root of unity, then ${\left( {1 + \omega  - {\omega ^2}} \right)^9}$  equals to?

 To solve the question, add and subtract ${\omega ^2}$ from the equation and then use $1 + \omega  + {\omega ^2} = 0$, to get the answer.

 We have ${\left( {1 + \omega  - {\omega ^2}} \right)^9}$ $ \\ $ This can be written as: ${\left( {1 + \omega  + {\omega ^2} - 2{\omega ^2}} \right)^9} $ $ \\ 1 + \omega  + {\omega ^2} = 0 \\ {\left( { - 2{\omega ^2}} \right)^9} \\   \Rightarrow  - 512{\omega ^{18}} \\   \Rightarrow  - 512{\omega ^2}{\omega ^{16}} \\   \Rightarrow  - 512{\omega ^2} \\  $                              

 Solve to find the modulus and argument of $z = 1 + i\sqrt 3 $ and write the answer.

 Substitute the values with $r\sin \theta $ and $r\cos \theta $, then use ${\cos ^2}\theta  + {\sin ^2}\theta  = 1$ after equating the values.

We have $z = 1 + i\sqrt 3 $ $ \\ $ Let $r\cos \theta  = 1$ and $r\sin \theta  = \sqrt 3 $ $ \\ $ Now, on squaring and adding, we get $ \\ {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2} = {\left( 1 \right)^2} + {\left( {\sqrt 3 } \right)^2} \\   \Rightarrow {r^2}\left( {{{\cos }^2}\theta  + {{\sin }^2}\theta } \right) = 1 + 3 \\  $ We know, ${\cos ^2}\theta  + {\sin ^2}\theta  = 1$ $ \\ {r^2} = 4 \\   \Rightarrow r = \sqrt 4  \\   \Rightarrow r = 2 \\  $ Therefore, modulus $ = 2$ Now, $ \\  2\cos \theta  = 1 \\  2\sin \theta  = \sqrt 3  \\ \Rightarrow \cos \theta  = \dfrac{1}{2} \\  \sin \theta  = \dfrac{{\sqrt 3 }}{2} \\  $ Since both the values are negative and $\sin \theta $ and $\cos \theta $ are negative in the third quadrant, we get; $ \\ $ Argument $ =  - \left( {\pi  + \dfrac{\pi }{3}} \right)  =  - \dfrac{{4\pi }}{3}$

 Arun bought some vegetables at the rate of Rs $5$ per onion. He then purchases an equal number of potatoes at Rs  $2$ per potato and makes a profit of $20\% $ on onions and $15\% $ on potatoes. On selling all the vegetables, he earned a profit of Rs$390$. Then find the number of onions.

 Assume the amount of onions and potatoes you'll need. Then find out how much money you can make on each vegetable. By combining the individual profits and equating them with the overall benefit, you will determine the quantity of onions.

Let the number of onions bought by him be $x$. $ \\ $  Let the number of potatoes bought by Arun be $y$. $ \\ $ Cost of onions: Rs $5x$ $ \\ $ Cost of potatoes: Rs $2x$ So, $ \\ $ Profit earned on onions: $\dfrac{{20}}{{100}} \times 5x  = x$ $ \\ $ Profit earned on potatoes: $\dfrac{{15}}{{100}} \times 2x  = \dfrac{3}{{10}}x$ $ \\ $ As per the condition: $ \\ x + \dfrac{3}{{10}}x = 390 \\   \Rightarrow \dfrac{{10x + 3x}}{{10}} = 390 \\   \Rightarrow \dfrac{{13x}}{{10}} = 390 \\   \Rightarrow x = 300 \\ $ So, total number of onions $ = 300$