Complex Numbers and Quadratic Equations Test 0

We know that

$$|\dfrac{z_{1}}{z_{2}}|$$ $$=\dfrac{|z_{1}|}{|z_{2}|}$$
Hence

$$|\dfrac{\alpha+i\beta}{\beta+i\alpha}|$$

$$=\dfrac{|\alpha+i\beta|}{|\beta+i\alpha|}$$

$$=\dfrac{\sqrt{\alpha^{2}+\beta^{2}}}{\sqrt{\beta^{2}+\alpha^{2}}}$$

$$=1$$.

$$m_{1}=|1+4i|=\sqrt{1+16}=\sqrt{17}$$.
$$m_{2}=|3+i|=\sqrt{9+1}=\sqrt{10}$$
$$m_{3}=|1-i|=\sqrt{1+1}=\sqrt{2}$$
$$m_{4}=|2-3i|=\sqrt{4+9}=\sqrt{13}$$
Hence
$$m_{1}>m_{4}>m_{2}>m_{3}$$

Given, $$z=-3+3i$$

The three vertices are $$A (1,1) , B (1,-1) $$ and $$C (0,2) $$

$$z_{1}=a_{1}+ib_{1}; z_{2}=a_{2}+ib_{2}$$
$$z_{1}z_{2}=(a_{1}+ib_{1})(a_{2}+ib_{2})$$
$$=a_{1}a_{2}+(a_{1}b_{2})i+(a_{2}b_{1})i-b_{1}b_{2}$$
$$=a_{1}a_{2}-b_{1}b_{2}+(a_{1}b_{2}+a_{2}b_{1})i$$
$$\therefore Re(z_{1}z_{2})=a_{1}a_{2}-b_{1}b_{2}$$
$$=Re(z_{1})\cdot Re(z_{2})-Im(z_{1})\cdot Im(z_{2})$$

Let $$z = x +iy $$
As $$z$$ is in the fourth quadrant, $$x >0 $$ and $$ y < 0 $$
$$\overline{z} = x - iy $$
In this case, the real part would be positive and the imaginary part would be positive. Hence, this would lie in the first quadrant. 

 $$-z  = -x - iy$$
 In this case, the real part would be negative and the imaginary part would be positive.  Hence this would lie in the second quadrant. 

 $$\overline{-z} = -x+iy $$
 In this case, the real part would be negative and the imaginary part would also be negative. Hence, this would lie in the third quadrant.  

The given series is a G.P with a common ratio of $$(1+i)$$.
Hence
 $$1+(1+i)+(1+i)^{2}+(1+i)^{3}$$

$$=\dfrac{1-(1+i)^{4}}{1-(1+i)}$$

Now
$$1+i=\sqrt{2}(\dfrac{1+i}{\sqrt{2}})$$

$$=\sqrt{2}(e^{i\frac{\pi}{4}})$$
Hence
$$\dfrac{1-(1+i)^{4}}{-i}$$

$$=\dfrac{1-(\sqrt{2}(e^{i\frac{\pi}{4}}))^{4}}{-i}$$

$$=\dfrac{1-4(e^{i\pi})}{-i}$$

$$=\dfrac{1-4(-1)}{-i}$$

$$=\dfrac{5}{-i}$$

$$=5i$$

$$(x+iy)(2+cos\theta+isin\theta)=3$$

$$2x+x(cos\theta+isin\theta)+i2y+iy(cos\theta+isin\theta)=3$$

$$2x+xcos\theta-ysin\theta+i(xsin\theta+2y+ycos\theta)=3$$

Therefore, equating the real and imaginary parts of LHS and RHS gives us

Let
$$z_{1}=x_{1}+iy_{1}$$
$$z_{2}=x_{2}+iy_{2}$$

$$z_{1}+z_{2}=(x_{1}+x_{2})+i(y_{1}+y_{2})$$ ...(i)

Now
$$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$$

$$\sqrt{(x_{1}+x_{2})^{2}+(y_{1}+y_{2})^{2}}=\sqrt{x_{1}^{2}+y_{1}^{2}}+\sqrt{x_{2}^{2}+y_{2}^{2}}$$

Squaring both sides give us
$$(x_{1}+x_{2})^{2}+(y_{1}+y_{2})^{2}=(x_{1})^{2}+(x_{2})^{2}+(y_{1})^{2}+(y_{2})^{2}+2\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}$$

$$(x_{1})^{2}+(x_{2})^{2}+(y_{1})^{2}+(y_{2})^{2}+2x_{1}x_{2}+2y_{1}y_{2}=(x_{1})^{2}+(x_{2})^{2}+(y_{1})^{2}+(y_{2})^{2}+2\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}$$

$$2x_{1}x_{2}+2y_{1}y_{2}=2\sqrt{x_{1}^{2}+y_{1}^{2}}\sqrt{x_{2}^{2}+y_{2}^{2}}$$

$$(x_{1}x_{2}+y_{1}y_{2})^{2}=x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}$$
$$x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+2x_{1}x_{2}y_{1}y_{2}=x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}$$
$$2x_{1}x_{2}y_{1}y_{2}=x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}$$
Or
$$x_{1}^{2}y_{2}^{2}+y_{1}^{2}x_{2}^{2}-2x_{1}x_{2}y_{1}y_{2}=0$$
Or
$$(x_{1}y_{2}-x_{2}y_{1})^{2}=0$$
Or
$$x_{1}y_{2}-x_{2}y_{1}=0$$
$$x_{1}y_{2}=x_{2}y_{1}$$
Or
$$\dfrac{x_{1}}{y_{1}}=\dfrac{x_{2}}{y_{2}}$$

Hence $$z_{1}$$  and $$z_{2}$$ are colinear.