Can we always use Euler's formula for complex numbers?

I was thinking of this today as I was looking over my complex analysis notes.

If you have some complex number $z$, then we can define it using Euler's formula as $z=a+ib=\cos\theta+i \sin\theta$. Say we have the case that $z=3+4i=25(\cos\theta+i\sin\theta)$. Then $25\cos \theta=3$, and $25\sin\theta=4$. But this would mean that

$$\theta=\cos^{-1}\left(\frac{3}{25}\right) =\sin^{-1}\left(\frac{4}{25}\right).$$

How can this be true if $\cos^{-1}\left(\frac{3}{25}\right)=83.107 \text{ degrees}$ and $\sin^{-1}\left(\frac{4}{25}\right)=9.206 \text{ degrees}$? Does this mean that we can only have certain values of $z$ in order to use Euler's formula ?

For $z=3+4i$; $$r=\sqrt{3^2+4^2}=5$$ $$\theta=\tan^{-1}(\frac{4}{3})\approx0.9273^c$$

Hence $z=5(\cos[0.9273]+i\sin[0.9273])$

Note via use of the $3,4,5$ triangle, we can tell that $\cos\theta=\frac{3}{5}$ and $\sin\theta=\frac{4}{5}$.

You simply missed the fact that you need to square root for $r$.

In the representation $a+bi=r(\cos(\theta)+i\sin(\theta))$, we have that $r=\sqrt{a^2+b^2}$. So in the case where $a=3$ and $b=4$; we have $r=\sqrt{3^2+4^2}=\sqrt{25}=5$ and not $25$.

• Ahhh now it works out , I just took my notes down wrong, silly me. Thank God though, if I had been right about what I thought i noticed in my notes I would have been far more confused XD. – excalibirr May 8 '18 at 21:18
• glad to know we got there in the end! no worries though; we all have those moments. – thesmallprint May 8 '18 at 21:32

You're misrepresenting Euler's formula. It is in fact $z=re^{i\theta} = r(\cos(\theta)+i\sin(\theta))$, where $r:=|z|=\sqrt{a^2+b^2}$, so to get the angles, first normalize $z$ via $z/|z|$.

In the first place, no, we don't always have $a+bi=\cos\theta+i\sin\theta$. WHat we have is$$a+bi=\sqrt{a^2+b^2}(\cos\theta+i\sin\theta).$$

On the other hand, both $\arccos\left(\frac35\right)$ and $\arcsin\left(\frac45\right)$ are equal to about $53.1301^\circ$.