# Help understanding conversion to polar form

I'm reading Spivak's Calculus, the chapter about complex numbers. It describes how complex numbers are converted from cartesian form to polar form:

For any complex number $$z\ne 0$$ we can write

$$z=|z| \frac{z}{|z|}$$

in this expression, $$|z|$$ is a positive real number, while:

$$\left|\frac{z}{|z|}\right| = \frac{|z|}{|z|} = 1$$

so that $$z/|z|$$ is a complex number of absolute value $$1$$. Now any complex number $$a = x + iy$$ with $$1 = |a| = x^2 + y^2$$ can be written in the form

$$a = (\cos\theta, \sin\theta) = \cos(\theta) + i\sin(\theta)$$

I understand in general the polar form of complex numbers. However, in the description above I don't understand why we first write

$$z=|z| \frac{z}{|z|}$$

And then conclude from there that a complex number with absolute value $$1$$ can be written in the form $$a = (\cos\theta, \sin\theta)$$.

Can anyone help filling in the gaps.

• $|z|$ is a magnitude, while $\frac{z}{|z|}$ is in a sense a direction – Henry Feb 28 '19 at 0:14
• @Henry I see that makes sense, but how does it help in deriving the polar form then? I mean, how does it relate to $\cos\theta, \sin\theta$ then? – Max Feb 28 '19 at 0:30

The thing to understand is that the set of all points $$(a,b)$$ where $$a^2 + b^2 = 1$$ are precisely the points $$(a,b)$$ on the unit circle and they are in one to one correspondence with the real numbers $$[0, 2\pi)$$ so that $$(a,b)=(\cos \theta, \sin \theta) \leftrightarrow m\angle AOX; A=(a,b); O=(0,0); X=(1,0)$$. That should be absolute basic first week of trigonometry.

Now if $$z = m+ni$$ then $$|z|= \sqrt{z\cdot \overline z} = (a+bi)(a-bi)=\sqrt{m^2 + n^2} =1$$.

Now, claiming the trivial fact that for $$z \ne 0$$ we have $$|z|\frac z{|z|} = z$$, serves to draw attention that there is a complex number $$w = \frac {z}{|z|} = \frac m{\sqrt{m^2 + n^2}} + i\frac n{\sqrt{m^2 + n^2}} = a+bi$$ where $$a = m{\sqrt{m^2 + n^2}}$$ and $$b = \frac n{\sqrt{m^2 + n^2}}$$ where $$|w| = 1$$ and $$a^2 + b^2 = 1$$.

Such a complex number will always exist and $$|w|$$ will equal $$1$$ and if $$w = a+bi$$ then.... $$|w| = \sqrt {a^2 + b^2} =1$$ and therefor $$a^2 + b^2 = 1$$.

And by basic trig, there cooresponds a precise and unique $$\theta: 0 \le \theta < 2\pi$$ so that $$a = \cos \theta$$ and $$b = sin \theta$$.

This isn't profound. When I was in high school, I would have been taught that if we plotted the number $$z=m+ni\ne 0$$ on the Complex Plane as the point $$(a,b)$$ then the distance from $$(a,b)$$ to $$(0,0)$$ would be $$r= \sqrt{a^2 + b^2}$$ and the angle of $$(a,b)$$ to $$(0,0)$$ to the $$x-axis$$ would be some precise angle $$\theta$$, and that of this $$r > 0$$ and $$\theta$$ than $$(a,b)$$ would be the only such point and that $$a+bi$$ is the unique one and only complex number with that magnitude and with that angle.

And that's all Spivak is saying.

But what's wrong with saying that. Well, that presupposes that the "Complex Plane" makes sense and has been defined and that we can talk about complex numbers $$a+bi$$ as points in the plane. This is the argument that we can.

Because if $$a=\cos\theta+i\sin\theta \tag{1}$$ then $$|a|=\cos^2\theta+\sin^2\theta=1\tag{2}$$ by the high school formula.

Effectively what this is saying is that we can write $$z_1=|z_1|z_2\tag{3}$$ then taking the modulus of both sides it becomes obvious that $$|z_2|=1$$ (Or, as the book does, divide by $$|z_1|$$, but to me taking the modulus makes it more obvious)

I showed at the start that if $$|z_2|=1$$ then $$z_2=\cos\theta+i\sin\theta$$ for some $$\theta$$.

It then follows that $$(3)$$ can be rewritten as: $$z_1=|z_1|(\cos\theta+i\sin\theta)$$ which is standard polar form.