Answer
Verified
91.5k+ views
Hint: We can use the conjugate multiplication of the denominator in both numerator and
denominator and hence simplify the complex number and find the real and imaginary part of the
complex number and check for the quadrant.
Complete step-by-step answer:
A complex number has two parts, real and imaginary. These represents the x and y coordinates of
the point being represented by the complex number. Here, we have been given with two complex
numbers
${{z}_{1}}=\sqrt{3}+\iota \sqrt{3}............(i)$
and ${{z}_{2}}=\sqrt{3}+\iota .......(ii)$
Calculating \[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}\] by substituting the values from (i) and
(ii)
$\Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left( \dfrac{\sqrt{3}+\iota
\sqrt{3}}{\sqrt{3}+\iota } \right)}^{50}}$
Multiplying and dividing by the conjugate of the ${{z}_{2}}$ in numerator and denominator and then
solving, we get,
\[\begin{align}
& \Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left( \dfrac{\sqrt{3}+\iota
\sqrt{3}}{\sqrt{3}+\iota }\times \dfrac{\sqrt{3}-\iota }{\sqrt{3}-\iota } \right)}^{50}} \\
& \Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{\left( \sqrt{3}+\iota
\sqrt{3} \right)\left( \sqrt{3}-\iota \right)}{\left( \sqrt{3}+\iota \right)\left( \sqrt{3}-\iota \right)}
\right]}^{50}} \\
\end{align}\]
Now on further simplification by multiplication we get,
\[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{\left( \sqrt{3} \right)\left( \sqrt{3}
\right)-\iota \sqrt{3}+\left( \iota \sqrt{3} \right)\left( \sqrt{3} \right)-\iota \left( \iota \sqrt{3}
\right)}{\left( \sqrt{3} \right)\left( \sqrt{3} \right)+\iota \left( \sqrt{3} \right)-\iota \left( \sqrt{3}
\right)-\iota \left( \iota \right)} \right]}^{50}}..................(iii)\]
As we know ${{\iota }^{2}}=-1$, substituting in (iii)
\[\Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{3-\iota \sqrt{3}+3\iota
+\sqrt{3}}{3+1} \right]}^{50}}\]
\[\Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{\left( 3+\sqrt{3}
\right)+\iota \left( 3-\sqrt{3} \right)}{4} \right]}^{50}}\]
Now,
$\dfrac{3+\sqrt{3}}{4}\succ 0$ and $\dfrac{3-\sqrt{3}}{4}\succ 0$
Both the real and imaginary parts are positive and this means the point represented by the complex number lies in the first quadrant as x and y coordinates of a point in the first quadrant are always positive.
Hence, the complex number \[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}\]is in first quadrant.
Final answer is option (a).
Note: The chances of mistakes are if conjugate of denominator is not multiplied correctly or the sign of the conjugate in numerator or denominator is altered mistakenly or ${{\iota }^{2}}=-1$ is substituted
wrongly as 1. The caution must be taken for interpreting the point’s position in the quadrant.
denominator and hence simplify the complex number and find the real and imaginary part of the
complex number and check for the quadrant.
Complete step-by-step answer:
A complex number has two parts, real and imaginary. These represents the x and y coordinates of
the point being represented by the complex number. Here, we have been given with two complex
numbers
${{z}_{1}}=\sqrt{3}+\iota \sqrt{3}............(i)$
and ${{z}_{2}}=\sqrt{3}+\iota .......(ii)$
Calculating \[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}\] by substituting the values from (i) and
(ii)
$\Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left( \dfrac{\sqrt{3}+\iota
\sqrt{3}}{\sqrt{3}+\iota } \right)}^{50}}$
Multiplying and dividing by the conjugate of the ${{z}_{2}}$ in numerator and denominator and then
solving, we get,
\[\begin{align}
& \Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left( \dfrac{\sqrt{3}+\iota
\sqrt{3}}{\sqrt{3}+\iota }\times \dfrac{\sqrt{3}-\iota }{\sqrt{3}-\iota } \right)}^{50}} \\
& \Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{\left( \sqrt{3}+\iota
\sqrt{3} \right)\left( \sqrt{3}-\iota \right)}{\left( \sqrt{3}+\iota \right)\left( \sqrt{3}-\iota \right)}
\right]}^{50}} \\
\end{align}\]
Now on further simplification by multiplication we get,
\[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{\left( \sqrt{3} \right)\left( \sqrt{3}
\right)-\iota \sqrt{3}+\left( \iota \sqrt{3} \right)\left( \sqrt{3} \right)-\iota \left( \iota \sqrt{3}
\right)}{\left( \sqrt{3} \right)\left( \sqrt{3} \right)+\iota \left( \sqrt{3} \right)-\iota \left( \sqrt{3}
\right)-\iota \left( \iota \right)} \right]}^{50}}..................(iii)\]
As we know ${{\iota }^{2}}=-1$, substituting in (iii)
\[\Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{3-\iota \sqrt{3}+3\iota
+\sqrt{3}}{3+1} \right]}^{50}}\]
\[\Rightarrow {{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}={{\left[ \dfrac{\left( 3+\sqrt{3}
\right)+\iota \left( 3-\sqrt{3} \right)}{4} \right]}^{50}}\]
Now,
$\dfrac{3+\sqrt{3}}{4}\succ 0$ and $\dfrac{3-\sqrt{3}}{4}\succ 0$
Both the real and imaginary parts are positive and this means the point represented by the complex number lies in the first quadrant as x and y coordinates of a point in the first quadrant are always positive.
Hence, the complex number \[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}\]is in first quadrant.
Final answer is option (a).
Note: The chances of mistakes are if conjugate of denominator is not multiplied correctly or the sign of the conjugate in numerator or denominator is altered mistakenly or ${{\iota }^{2}}=-1$ is substituted
wrongly as 1. The caution must be taken for interpreting the point’s position in the quadrant.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If the distance between 1st crest and the third crest class 11 physics JEE_Main
A man of mass 50kg is standing on a 100kg plank kept class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main