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If $z_1$ and $z_2$ both satisfy $z+\bar{z}=2|z-1|$, $\arg(z_1-z_2)=\dfrac{\pi}{4}$, then find $\Im(z_1+z_2)$

My Attempt $$ z_1+\bar{z}_1=2|z_1-1|\quad\&\quad z_2+\bar{z}_2=2|z_2-1|\quad\&\quad \arg(z_1-z_2)=\dfrac{\pi}{4}\\ z_1-z_2=\frac{|z_1-z_2|}{\sqrt{2}}\Big(1+i\Big)\\ \Re(z_1+z_2)=\frac{z_1+z_2+\bar{z}_1+\bar{z}_2}{2}=|z_1-1|+|z_2-1|\\ \Im(z_1+z_2)=\frac{z_1+z_2-\bar{z}_1-\bar{z}_2}{2i}\\ \Re(z_1-z_2)=\frac{z_1-z_2+\bar{z}_1-\bar{z}_2}{2}=\frac{|z_1-z_2|}{\sqrt{2}}={|z_1-1|-|z_2-1|}\\ \Im(z_1-z_2)=\frac{z_1-z_2-\bar{z}_1+\bar{z}_2}{2i}=\frac{|z_1-z_2|}{\sqrt{2}} $$ My reference gives the solution $2$, but how do I find the solution without actually representing $z=a+ib$ ?

Sooraj S
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2 Answers2

2

It's easy if you consider

$z_1=a_1+ib_1$ , $z_2=a_2+ib_2$

$\bar{z_1}=a_1-ib_1$ , $\bar{z_2}=a_2-ib_2$

Given $z+\bar{z}=2|z-1|$

Since, $z_1$ and $z_2$ both satisfy, $z+\bar{z}=2|z-1|$, we can say that $$z_1+\bar{z_1}=2|z_1-1|$$ and $$z_2+\bar{z_2}=2|z_2-1|$$

Firstly, $$z_1+\bar{z_1}=2|z_1-1|$$ $$a_1+ib_1+a_1-ib_1=2|(a_1-1)+ib_1|$$ After solving we get, $$b_1^2=2a_1-1\tag1$$ Similarly, we get $$b_1^2=2a_2-1\tag3$$

Now, $$arg(z_1-z_2)=\dfrac{\pi}{4}$$ $$arg(a_1+ib_1-a_2-ib_2)=\dfrac{\pi}{4}$$ $$\dfrac{b_1-b_2}{a_1-a_2}=\tan\dfrac{\pi}{4}$$ $$\dfrac{b_1-b_2}{a_1-a_2}=1$$ $$b_1-b_2=a_1-a_2\tag3$$ Now use the equations $(1),(2)$ and $(3)$ to get the final answer.

Key Flex
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Without representing $z=a+bi$. \begin{align} z+\bar z&=2|z-1|\\ (z+\bar z)^2&=4|z-1|^2=4(z-1)(\bar z-1)\\ z^2+2z\bar z+\bar z^2&=4(z\bar z-z-\bar z+1)\\ z^2-2z\bar z+\bar z^2&=4-4(z+\bar z)\\ (z-\bar z)^2&=4-4(z+\bar z)\tag{1} \end{align} Now take (1) for $z_2$ minus (1) for $z_1$ and use $$ z_1-z_2=\frac{|z_1-z_2|}{\sqrt{2}}(1+i)\tag{2} $$ to get \begin{align} (z_2-\bar z_2)^2-(z_1-\bar z_1)^2&=4(z_1+\bar z_1-z_2-\bar z_2)\\ (\underbrace{z_2-\bar z_2-z_1+\bar z_1}_{-\sqrt2i|z_1-z_2|})(\underbrace{z_2-\bar z_2+z_1-\bar z_1}_{2i\operatorname{Im}(z_1+z_2)})&=4(\underbrace{z_1+\bar z_1-z_2-\bar z_2}_{\sqrt2|z_1-z_2|})\\ |z_1-z_2|(\operatorname{Im}(z_1+z_2)-2)&=0. \end{align}

A.Γ.
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