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Are there any complex expressions that cannot be simplified to the form $a+jb$, where a and b are real numbers?

For example, $$\frac{1}{j}=0+j(-1),\hspace{0.5cm}e^j=\cos(1)+j\sin(1),\hspace{0.5cm}\sin(j)=0+j\frac{e^2-1}{2e}$$

From what I understand, all complex numbers must exist somewhere on the complex plane where a and b are the coordinates. But some expressions don't have any obvious way to be simplified: $$\ln(1+j)=???,\hspace{0.5cm}\arctan(j)=???$$ If every expression can be simplified, are there any good references or list of identities?

msm
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Dan
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  • Yes all complex numbers can be written in the form $a+ib$ for $a,b\in\mathbb{R}$. But this is not enough to define complex numbers, there are a few more restrictions. – StubbornAtom Oct 05 '16 at 06:23

5 Answers5

1

To get $\ln(z)$ for any complex $z$ (where $z \ne 0$), write $z = |z|y$.

Then $|y| = 1$, so there is a real $t$ such that $y = e^{it} =\cos(t)+i\sin(t) $, so $z = |z|y = |z|(\cos(t)+i\sin(t)) = |z|\cos(t)+i|z|\sin(t) $.

Since $|1+j| = \sqrt{2}$, $(1+j) =\sqrt{2}\frac{1+j}{2} =\sqrt{2}e^{i\pi/4} $, so $\ln(1+j) =\ln(\sqrt{2}e^{i\pi/4}) =\dfrac{\ln(2)}{2}+i\pi/4 $ (you can throw in $+2\pi i n$ if you want).

marty cohen
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Yes, all complex expressions that produce a complex number ($\arctan(i)$ doesn't) can be written in rectangular form. You can generally use a Taylor Series or convert your expression to another form to find the complex number in rectangular form. Here is a page that deals with a lot of these cases. Most of these results use $e^{ix}=\cos(x)+i\sin(x)$ to convert functions that don't seem to work with complex numbers into ones that do.

AlgorithmsX
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Assume $z=re^{i\theta}=x+iy$. Then $$\ln(z)=\ln(r)+\ln(e^{i\theta})=\ln(r)+i\theta=\ln(\sqrt{x^2+y^2})+i\arctan(\frac{y}{x})$$

Also regarding $\arctan(z)$, it is not difficult to show that $$\arctan(z)=\frac{1}{2i}\ln\left(\frac{i-z}{i+z}\right)$$

msm
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The first one is simple: $$\ln(1+j) = \ln(\sqrt{2} e^{j\pi/8}) = \frac12 \ln 2 + j\frac\pi8$$

For the second: if $$z = \tan w = \frac{\sin w}{\cos w} = \frac1j \frac{e^{jw} - e^{-jw}}{e^{jw}+e^{-jw}}$$ we can solve for $w = \arctan z$: $$ e^{2jw} = \frac{1+jz}{1-jz} $$ $$\arctan(z) = w = \frac12 \ln e^{2jw} = \frac12 \ln \left( \frac{1+jz}{1-jz} \right).$$ In particular when $z = j$: $$\arctan(j) = \frac12 \ln \left( \frac{1+j^2}{1-j^2} \right) = \frac12 \ln 0 $$ and, well, $\ln 0$ is undefined.

arkeet
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Any complex number $z$ can be expressed as $Re(z)+i Im(z)$.

yoyostein
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