# If ${{z}_{1}}=\sqrt{3}+\iota \sqrt{3}$ and ${{z}_{2}}=\sqrt{3}+\iota $ , then the complex number \[{{\left( \dfrac{{{z}_{1}}}{{{z}_{2}}} \right)}^{50}}\] lies in the

(a) First quadrant

(b) Second quadrant

(c) Third quadrant

(d) Fourth quadrant

Last updated date: 22nd Mar 2023

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Answer

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306k+ 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.

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