Question

$2z_4=?$

• A. $z_1(1+i\sqrt{3})+z_3(1-i\sqrt{3})$
• B. $z_1(1-i\sqrt{3})-z_3(1+i\sqrt{3})$
• C. $z_1(1-i\sqrt{3})+z_3(1+i\sqrt{3})$
• D. $z_1(1+i\sqrt{3})-z_3(1+i\sqrt{3})$

Answer 1

Answer: (C)

$z_1(1-i\sqrt{3})+z_3(1+i\sqrt{3})$

$z_1=(z_3-z_2)\times e^{i\frac{\pi }{3}}+z_2$

$\Rightarrow z_1=\frac{1}{2}\times (z_3-z_2)(1+i\sqrt{3})+z_2$

$\Rightarrow 2z_1=(z_3-z_2)(1+i\sqrt{3})+2z_2$

$\Rightarrow 2z_1=z_3(1+i\sqrt{3})+z_2(1-i\sqrt{3})$

$\Rightarrow z_2=\frac{1}{1-i\sqrt{3}}\times [2z_1-z_3(1+i\sqrt{3}\left)\right]$

$\Rightarrow 2z_2=\frac{1}{2}(2+i2\sqrt{3})z_1-\frac{1}{2}(1+i\sqrt{3}{)}^2{z}_3$

$\Rightarrow 2z_2=(1+i\sqrt{3})z_1+(1-i\sqrt{3})z_3$

Now, $2(z_2+z_4)=2(z_1+z_3)$ since the diagonals bisect each other.
$\therefore 2z_4=2z_1+2z_3-2z_2$

$\therefore 2z_4=2z_1+2z_3-(1+i\sqrt{3})z_1-(1-i\sqrt{3})z_3$

i.e. $2z_4=(1-i\sqrt{3})z_1+(1+i\sqrt{3})z_3$