Expressing equations in polar form and the answers in Cartesian form

gazzer

Folks. I am doing a maths module and of the 7 questions I have done I am stuck on the last one.

This is the question.

(1) Express the equation z^3 = -64i in polar form. Hence solve the equation, expressing your answers in polar form.

(2) Express yours answers to part (1) in Cartesian Form.

(3) Draw a diagram showing the solutions to part (1) in the complex plane

I am looking for any websites or video tutorials that would help me in working this question out so would appreciate it if any of you here knew of any places? The books I have are not helping with this question.

Thanks

Michael Collins

Any complex number can be expressed in two ways:

1) Cartesian form: [latex] z = a + i \cdot b [/latex]
2) Polar form: [latex] z = r \cdot \hbox{e}^{i \cdot \theta} [/latex]

In the Cartesian form, [latex] a [/latex] and [latex]b[/latex] are the real and imaginary coodinates respectively.

In the polar form we imagine a line segment from the origin, to the point [latex]z[/latex] in the complex plane.

The value [latex]r[/latex] is the length of this line segment.

By Pythagoras, [latex] |z| = r = \sqrt{a^2+b^2} [/latex].

The value [latex]\theta[/latex], known as [latex] \arg(z) [/latex], is the angle this line segment makes with the positive real axis. So [latex] \arg(z) = \theta [/latex] can be found using trigonometry. For example, if we are in the first quadrant (i.e. a > 0, b > 0), then

[latex]\arg(z) = \theta = \tan^{-1}(b/a) [/latex]

Note that you can add any multiple of [latex] 2\pi [/latex] to the angle [latex] \theta [/latex] without affecting the result (you will need this idea later).

For example, let's take

[latex] \displaystyle z = \frac{1}{\sqrt{2}} + i \cdot\frac{1}{\sqrt{2}} [/latex]

then

[latex] \displaystyle |z| = r = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} =\sqrt{1/2 + 1/2} = 1 [/latex]

and, since we are in the first quadrant

[latex] \displaystyle \arg(z) = \theta = \tan^{-1}(b/a) = \tan^{-1}(1) = \pi/4\hbox{ or } 45^{\circ}[/latex]

so, we can express [latex]z[/latex] in the polar form

[latex] z = r \cdot \hbox{e}^{i \cdot \theta} = 1 \cdot \hbox{e}^{i \cdot \pi/4} = \hbox{e}^{i \cdot \pi/4}[/latex]

Now have a go at putting your [latex] z^3 [/latex] in polar form, and write back what ya get.

gazzer

Thanks so much for taking the time to answer my query. What I am getting so far is

[latex] \displaystyle z = 0 + (-64)i[/latex]

then

[latex] \displaystyle |z| = r = \sqrt{0^2 + -64^2} =\sqrt{0 + 64} = 64 [/latex]

[latex] \displaystyle \arg(z) = \theta = \tan^{-1}(-64/0) = 0???? [/latex]

If you divide b by a you get -64/0 which gives 0. Or am I missing something totally obvious?

Michael Collins

gazzer wrote: »

Thanks so much for taking the time to answer my query. What I am getting so far is

[latex] \displaystyle z = 0 + (-64)i[/latex]

then

[latex] \displaystyle |z| = r = \sqrt{0^2 + -64^2} =\sqrt{0 + 64} = 64 [/latex]

[latex] \displaystyle \arg(z) = \theta = \tan^{-1}(-64/0) = 0???? [/latex]

If you divide b by a you get -64/0 which gives 0. Or am I missing something totally obvious?

Whoops! You are correct for the magnitude part but I probably should have explained a bit better regarding the angle. Just draw a diagram and the angle should be obvious, i.e. what is the angle between a line that goes straight down on the imaginary axis, and the real axis itself?

You can work it out via maths too:

[latex] \displaystyle \arg(z) = \theta = \tan^{-1}(-64/0) = \tan^{-1}(-\infty) [/latex],

so the question is what value of [latex] \theta [/latex] corresponds to a [latex] \tan \theta [/latex] equal to [latex] -\infty [/latex]?

The answer is

[latex] \displaystyle \tan{\theta} \to -\infty\hbox{ as } \theta \to \pi/2 [/latex]

which means

[latex] \tan^{-1}(-\infty)[/latex] is?