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I have a text book question to find the principal argument of

$$ z = {i \over -2-2i}. $$

I know formulas where we find using $$ \tan^{-1} {y \over x}$$

but I am kinda stuck here can somebody please help.

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3 Answers 3

Rewrite as

$$z = -\frac{1+i}{4}$$

Note that $\Re{z} = \Im{z} = -1/4$, so that the argument of $z$ lies in the third quadrant (same sign, both negative). Because the real and imaginary parts are equal,

$$\text{Arg}\,{z} = \frac{-3 \pi}{4}$$

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yup got that part now to find the Arg(z) what should i do? –  Kishan Thobhani Mar 15 '13 at 13:38
    
@RonGordon, Accroding to you, the principal argument of Arg(a+ia) will be $\frac\pi4$ for $a\ne0$ right? But the principal argument lies $\in( −\pi,\pi]$ en.wikipedia.org/wiki/… –  lab bhattacharjee Mar 15 '13 at 14:37
    
@labbhattacharjee: I see. This is confusing because the principal value of the arctan lies between $-\pi/2$ and $\pi/2$. I think I have had run-ins of this type here before. –  Ron Gordon Mar 15 '13 at 14:44
    
@RonGordon, $Arg(1+i)$ should not be equal to $Arg(-1-i),$ right? –  lab bhattacharjee Mar 15 '13 at 14:45
    
@labbhattacharjee: I am not disagreeing with you. I know that when I want the correct quadrant, I use atan2. I know of "principal value" from the use of Arctan, that's my issue. But if that's the accepted use of the word, so be it. –  Ron Gordon Mar 15 '13 at 14:47

$$z=\frac i{-2-2i}=\frac12\frac {-i}{1+i}=\frac12\frac {-i(1-i)}{(1+i)(1-i)}=\left(-\frac14\right)+i\left(-\frac14\right)$$

Using the definition of $\arctan2,$

the principal value of the Argument of $z$ (which lies in $(-\pi,\pi]$) will be $\arctan 1 -\pi=\frac\pi4-\pi=-\frac{3\pi}4$ (as $\frac{\left(-\frac14\right)}{\left(-\frac14\right)}=1$)


Alternatively, using this, $$Arg \left(\frac {z_1z_2\cdots}{w_1w_2\cdots}\right)=\sum Arg(z_i)-\sum Arg(w_i)\pmod {(-\pi,\pi]}$$

$$\text{So, }Arg \left(\frac i{-2-2i}\right)= Arg(i)-Arg(-1)-Arg(1+i)$$ $$=\frac\pi2-\pi-\frac\pi4\pmod {(-\pi,\pi]}=-\frac{3\pi}4$$

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$z = \cfrac{i}{ -2-2i}=\cfrac{i(-2+2i)}{ (-2-2i)(-2+2i)}=\cfrac{i(-2+2i)}{ (-2-2i)\overline{(-2-2i)}}=\cfrac{i(-2+2i)}{ \left|-2-2i\right|^2}=\cfrac{-2i-2}{ \left|-2-2i\right|^2}=\cfrac{1}{ \left|-2-2i\right|^2}(-2-2i)$

Since $\cfrac{1}{ \left|-2-2i\right|^2}\in \Bbb R_+^*$, the principal argument of $z$ is also the principal argument of $-2-2i$ which you should be able to find.

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