# Thread: Polar form of -1 + i.

1. ## Polar form of -1 + i.

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

2. 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?

3. Well $r=\sqrt{1^2 + 1^2}=\sqrt{2}$
Angle is $315$

4. r is $\sqrt{1^2+1^2} = \sqrt{2}$

While $\theta = \tan^{-1}\left(\frac{-1}{1}\right), \theta \in (-\pi, \pi)$

Then $z={e\sqrt{2}}^{(i*{-0.758})}$ ?

6. I get $\displaystyle z= \sqrt{2}e^{\frac{3\pi}{4}i}$

7. Originally Posted by Critter314
Write $z=i-1$ in form of ${re}^{iv}$
$\text{Arg}(-1+\imath)=\dfrac{3\pi}{4}$.

So $\sqrt{2}\exp\left(\dfrac{3\imath\pi}{4}\right).$

8. How do I find an argument?
315 degrees in radians is $\frac{7\pi}{4}$

9. 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 $a=-1~\&~b=1$.

10. 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.