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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Complex Number

BITSAT 2016 · Mathematics · Complex Number

If z_ 1 = 3 +i 3 and z_ 2 = 3 +i , then the complex number ( z_ 1...

If z1=3+i3z_{1}=\sqrt{3}+i \sqrt{3} and z2=3+iz_{2}=\sqrt{3}+i, then the complex number (z1z2)50\left(\frac{\mathrm{z}_{1}}{\mathrm{z}_{2}}\right)^{50} lies in the :

Options

  1. Afirst quadrant
  2. Bsecond quadrant
  3. Cthird quadrant
  4. Dfourth quadrant

Correct answer

A. first quadrant

Step-by-step solution

(z1z2)50=(3+i33+i)50=\left(\frac{z_{1}}{z_{2}}\right)^{50}=\left(\frac{\sqrt{3}+i \sqrt{3}}{\sqrt{3}+i}\right)^{50}=\left[\left(\frac{\sqrt{3}(1+i)}{\sqrt{3}+i}\right)^{2}\right]^{25}=\left[\frac{3(2 i)}{3-1+2 \sqrt{3} i}\right]^{25}=\left(\frac{3 i}{1+\sqrt{3} i}\right)^{25}=\frac{3^{25} i^{25}}{\left(-2 \omega^{2}\right)^{25}} =−i.ω(32)25=−i(−1+3i2)(32)25=(32)25(32+12i)\begin{array}{l} =-i . \omega\left(\frac{3}{2}\right)^{25}=-\mathrm{i}\left(\frac{-1+\sqrt{3} \mathrm{i}}{2}\right)\left(\frac{3}{2}\right)^{25}\\ =\left(\frac{3}{2}\right)^{25}\left(\frac{\sqrt{3}}{2}+\frac{1}{2} \mathrm{i}\right) \end{array}Hence,(z1z2)50Hence, \left(\frac{z_{1}}{z_{2}}\right)^{50} lies in the first quadrant as both real and imaginary parts of this number are positive.
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