Courses
Courses for Kids
Free study material
Free LIVE classes
More

# 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
Total views: 306k
Views today: 8.86k
Verified
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.

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.

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