Write $\displaystyle z=i-1$ in form of $\displaystyle {re}^{iv}$

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- Mar 22nd 2011, 12:11 PMCritter314Polar form of -1 + i.
Write $\displaystyle z=i-1$ in form of $\displaystyle {re}^{iv}$

- Mar 22nd 2011, 12:18 PMHallsofIvy
It's a math exercise! And it is asking you to change a complex number written in Cartesian form to its "polar form". The simplest way to do that is to graph i- 1 as (-1, 1) (using the x-axis as the real axis, the y-axis as the imaginary axis). Now, your "r" is the distance from (0, 0) to (-1, 1) and your "v" is the angle the line from (0,0) to (-1, 1) makes with the positive real (x) axis. Can you calculate those?

- Mar 22nd 2011, 12:25 PMCritter314
Well $\displaystyle r=\sqrt{1^2 + 1^2}=\sqrt{2}$

Angle is $\displaystyle 315$ - Mar 22nd 2011, 12:28 PMpickslides
r is $\displaystyle \sqrt{1^2+1^2} = \sqrt{2}$

While $\displaystyle \theta = \tan^{-1}\left(\frac{-1}{1}\right), \theta \in (-\pi, \pi)$ - Mar 22nd 2011, 12:36 PMCritter314
I should type answer in radians?

Then $\displaystyle z={e\sqrt{2}}^{(i*{-0.758})}$ ? - Mar 22nd 2011, 12:40 PMpickslides
I get $\displaystyle \displaystyle z= \sqrt{2}e^{\frac{3\pi}{4}i} $

- Mar 22nd 2011, 12:41 PMPlato
- Mar 22nd 2011, 01:05 PMCritter314
How do I find an argument?

315 degrees in radians is $\displaystyle \frac{7\pi}{4}$ - Mar 22nd 2011, 01:15 PMPlato
The principle value of the argument of a complex number $\displaystyle z=a+bi$ not on any axis is found by the following.

$\displaystyle Arg(z) = \left\{ {\begin{array}{rl}

{\arctan \left( {\frac{b}

{a}} \right),} & {a > 0} \\

{\arctan \left( {\frac{b}

{a}} \right) + \pi ,} & {a < 0\;\& \,b > 0} \\ \\

{\arctan \left( {\frac{b}

{a}} \right) - \pi ,} & {a < 0\;\& \,b < \pi } \\

\end{array} } \right.$

Please note that $\displaystyle \mathif{i}-1=-1+\mathif{i}$ and there $\displaystyle a=-1~\&~b=1$. - Mar 22nd 2011, 02:12 PMHallsofIvy