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My solving so far :

$\arg(z_1z_2) = \pi/2 $ $\implies \arg(z_1) = \pi/2 - \arg(z_2) $

Let $z_2 = \theta ; |z_1| = r , |z_2| = r' $

$\implies z_2 = r'\{\cos\theta + i\sin\theta\}$ & $z_1 = r\{sin\theta + i\cos\theta\}$

As you can see r & r' do not cancel , so theres no way I can find $\theta$ through this approach .

We are so far taught about genral form , polar/trig.form , Euler form & Demiovre's Theroem .

Please answer accordingly .

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  • $\begingroup$ You have not yet used the given info that (1-i) Z1 =Z2 $\endgroup$ – user268307 Dec 17 '15 at 5:27
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Use this rule (which you should know):

$\arg(zw) = \arg(z) + \arg(w)$

So the first equation yields $-\frac{\pi}{4} + \arg(z_1) = \arg(z_2)$

And the second equation yields $\arg(z_1) + \arg(z_2) = \frac{\pi}{2}$

Solving those simultaneously very quickly gives you $\arg(z_2) = \frac{\pi}{8}$

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  • $\begingroup$ The first equation will yield $-\pi/4 + \arg(z_1) = \arg(2z_2 ) $ $\endgroup$ – Ricky Dec 17 '15 at 5:41
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    $\begingroup$ @Ricky Where did the coefficient of $2$ on the RHS come from? In any case, $\arg(kz) = \arg(z), k \in \mathbb{R^+} $. $\endgroup$ – Deepak Dec 17 '15 at 5:44
  • $\begingroup$ Missed including co.eff 2 , which was in actual question , Anyway , its the same story. Thanks $\endgroup$ – Ricky Dec 17 '15 at 5:45
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$$z_1=x+iy, z_2=a+ib\implies z_1z_2=xa-yb+i(bx+ay)$$

arg$(z_1z_2)=\dfrac\pi2\implies xa-yb=0\iff\dfrac xb=\dfrac ya=k$(say)

$\implies x=bk,y=ak$

Now $z_1(1-i)=z_2\iff a+ib=k(b+ai)(1-i)=(b+a)k+ik(a-b)$

$\implies a=(b+a)k\iff a(1-k)-bk=0$

and $b=k(a-b)\iff ak-b(k+1)=0$

Solve for $a,b$

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Write $1-i = \sqrt{2} e^{- i \pi/4}$, and let $z_1 = r_1 e^{i \theta_1}, z_2 = r_2 e^{i \theta_2}$ with $\theta_1, \theta_2 \in [0, 2 \pi)$.

Then, $(1-i) z_1 = z_2$ is $\sqrt{2} e^{-i \pi/4} r_1 e^{i \theta_1} = r_2 e^{i \theta_2}$ or equivalently, $\sqrt{2} r_1 e^{i (\theta_1 - \pi/4)} = r_2 e^{i \theta_2}$.

Thus, you must have $\theta_1 - \pi/4 = \theta_2 + 2 \pi k$ for some integer $k$.

Note that $z_1z_2 = r_1 r_2 e^{i (\theta_1 + \theta_2)}$ so $arg(z_1 z_2) = \theta_1 + \theta_2= \pi/2$.

Now combine these statements.

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