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Trying to understand how to express this complex number $\ln\left(\frac{1}{1+j}\right)$ into rectangular form.

I thought to use $x = r\cos(\theta)$ and $ y = r\sin(\theta)$, but I am pretty sure that is wrong.

Geno C
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  • There was absolutely no reason to edit the question, it is my preference to decide if I want parenthesis around theta. So irrelevant and trivial to edit the question. – Geno C Feb 01 '21 at 07:41
  • Can you calculate the magnitude and the argument of $1/(1+j)$ (that is, can you express $1/(1+j)$ in polar form)? If so, do you know how to compute the complex logarithm of a number in that form? – Greg Martin Feb 01 '21 at 07:43
  • @GenoC The edited version corrected typesetting. – Ken Feb 01 '21 at 07:46
  • @GregMartin, I am so stupid, I am sorry I see what you mean now. – Geno C Feb 01 '21 at 08:06
  • You're not stupid—you're simply learning! – Greg Martin Feb 01 '21 at 17:05

3 Answers3

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If by $\ln$, you mean the main branch of the complex logarithm (remember, the complex logarithm is a fairly complicated topic), then use the fact that $$\log z = \log|z| + i\cdot \mathrm{Arg}(z).$$

5xum
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$$\ln\frac1{1+i}=-\ln(1+i)=-\ln\sqrt2e^{i\pi/4}=-(\ln\sqrt2+i\pi/4)=\ln\frac1{\sqrt2}-i\pi/4$$ with the usual caveat about $\ln$'s multivaluedness.

Parcly Taxel
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$$\ln\Bigl(\frac{1}{1+i}\Bigr)=-\ln\Bigl(1+i\Bigr)= -\ln\Bigl(\sqrt{2}\Bigl(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\Bigr)\Bigr)$$

$$-\ln\Bigl(\sqrt{2}\Bigl(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\Bigr)\Bigr)=-\ln\Bigl(\sqrt{2}\Bigl(\cos\frac{\pi}{4}+{i}\sin\frac{\pi}{4}\Bigr)\Bigr)$$ $r(\cos\theta +i\sin\theta)$ can be written as $re^{i\theta}$

$$-\ln\Bigl(\sqrt{2}\Bigl(\cos\frac{\pi}{4}+{i}\sin\frac{\pi}{4}\Bigr)\Bigr)=-\ln\Bigl(\sqrt{2}\Bigl(e^{i\frac{\pi}{4}}\Bigr)\Bigr)=-\ln\sqrt{2} -\ln{e^{i\frac{\pi}{4}}}$$

$$= \ln\frac{1}{\sqrt{2}} -i\frac{\pi}{4}$$

Smriti Sivakumar
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