Complex Numbers, Polar Coordinates

**jschlarb** · June 22nd 2008, 03:39 PM

Okay, I have this so far

z= -1+7i

As far as what I have for polar coordinate equation, this is it:

z= 6.93(cos180+ i sin?)

My problem is that 7 cannot be converted into an angle.

What would I do for this?

**mr fantastic** · June 22nd 2008, 03:46 PM

Originally Posted by jschlarb

Okay, I have this so far

z= -1+7i

As far as what I have for polar coordinate equation, this is it:

z= 6.93(cos180+ i sin?)

My problem is that 7 cannot be converted into an angle.

What would I do for this?

Please show >all< of your working ..... Then it will be much easier to explain your mistakes to you ....

**jschlarb** · June 22nd 2008, 03:56 PM

z=-1+7i

a=-1, b=7, r= $[square root] a^2+b^2$ = 6.93 (subbed in -1 and 7)

$cos^-1 (-1) = 180, sin^-1 (7) = does not exist$

z=6.93(cos180+ i sin ?)

You'll have to forgive me, I'm not sure of the tag for square roots (if there is a link to all math signs, could you please link it?)

**Plato** · June 22nd 2008, 04:03 PM

Originally Posted by jschlarb

z= -1+7i or polar coordinates

First just do the basic computations: $\theta = \arg \left( { - 1 + 7i} \right) = \arctan \left( {\frac{7}{{ - 1}}} \right) + \pi \,\& \,\sqrt {\left( { - 1} \right)^2 + 7^2 } = 5\sqrt 2$ .
Now $z = 5\sqrt 2 e^{i\theta }$ .

**mr fantastic** · June 22nd 2008, 04:19 PM

Originally Posted by jschlarb

z=-1+7i

a=-1, b=7, r= $[square root] a^2+b^2$ = 6.93 (subbed in -1 and 7)

$cos^-1 (-1) = 180, sin^-1 (7) = does not exist$

z=6.93(cos180+ i sin ?)

You'll have to forgive me, I'm not sure of the tag for square roots (if there is a link to all math signs, could you please link it?)

You should know that the argument of z = a + ib is $\theta = \tan^{-1} \frac{b}{a}$ where you have to be careful what quadrant (and hence value) you choose for $\theta$ ..... A simple argand diagram showing z = a + ib will help you make that decision.

For the question you've posted, $\theta = \tan^{-1} \frac{7}{-1} = \tan^{-1} (-7)$ .

An argand diagram easily shows that z = -1 + 7i lies in the 2nd quadrant, so $\theta = \tan^{-1} (-7) + \pi \approx -1.429 + \pi = 1.713$ radians.

Therefore $z = -1 + 7i = r (\cos \theta + i \sin \theta) = .....$

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