# Polar form of -1 + i.

• Mar 22nd 2011, 12:11 PM
Critter314
Polar form of -1 + i.
Write $z=i-1$ in form of ${re}^{iv}$
• Mar 22nd 2011, 12:18 PM
HallsofIvy
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 PM
Critter314
Well $r=\sqrt{1^2 + 1^2}=\sqrt{2}$
Angle is $315$
• Mar 22nd 2011, 12:28 PM
pickslides
r is $\sqrt{1^2+1^2} = \sqrt{2}$

While $\theta = \tan^{-1}\left(\frac{-1}{1}\right), \theta \in (-\pi, \pi)$
• Mar 22nd 2011, 12:36 PM
Critter314
Then $z={e\sqrt{2}}^{(i*{-0.758})}$ ?
• Mar 22nd 2011, 12:40 PM
pickslides
I get $\displaystyle z= \sqrt{2}e^{\frac{3\pi}{4}i}$
• Mar 22nd 2011, 12:41 PM
Plato
Quote:

Originally Posted by Critter314
Write $z=i-1$ in form of ${re}^{iv}$

$\text{Arg}(-1+\imath)=\dfrac{3\pi}{4}$.

So $\sqrt{2}\exp\left(\dfrac{3\imath\pi}{4}\right).$
• Mar 22nd 2011, 01:05 PM
Critter314
How do I find an argument?
315 degrees in radians is $\frac{7\pi}{4}$
• Mar 22nd 2011, 01:15 PM
Plato
Quote:

Originally Posted by Critter314
How do I find an argument?

The principle value of the argument of a complex number $z=a+bi$ not on any axis is found by the following.
$Arg(z) = \left\{ {\begin{array}{rl}
{\arctan \left( {\frac{b}
{a}} \right),} & {a > 0} \\
{\arctan \left( {\frac{b}
{a}} \right) + \pi ,} & {a < 0\;\& \,b > 0} \\ \\
{\arctan \left( {\frac{b}
{a}} \right) - \pi ,} & {a < 0\;\& \,b < \pi } \\
\end{array} } \right.$

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

Originally Posted by Critter314
How do I find an argument?
315 degrees in radians is $\frac{7\pi}{4}$

315= 7(45) which is just 45 degrees short of the full circle. That would correspond to 1- i, not -1+ i.