This is for a Principles of Applied Mathematic college course.
I am having trouble expressing ln(-1-i) in the form a+ib
Thank you
What do you mean by "having trouble"? Have you tried anything at all on this? What do you understand about this? There is a general formula for ln(x+ iy) but I presume you do not know that. Do you known how to write a+ bi in "polar" or "exponential form, $\displaystyle re^{i\theta}$? If you write $\displaystyle a+ib$ in the form $\displaystyle re^{i\theta}$, then $\displaystyle ln(a+ bi)= ln(re^{i\theta})= ln(r)+ i\theta$.
(That is the "principle value". Since $\displaystyle \theta+ 2k\pi$ is the same as angle $\displaystyle \theta$ for any integer, k, the general formula is $\displaystyle ln(re^{i\theta})= ln(r)+ i(\theta+ 2k\pi)$ for any integer, k.)
that's the value closest to the origin, yes.
$\ln(z)$ is multivalued
you might want to fully simplify it to
$\dfrac 1 2 \ln(2) - i \dfrac {3\pi}{4}$
and then including the multiple values we can write
$\ln(-1-i) = \dfrac 1 2 \ln(2) - i\left(\dfrac{3\pi}{4} \pm 2\pi k\right),~~k \in \mathbb{Z}$
$\displaystyle \begin{align*} z &= \ln{ \left( -1 - \mathrm{i} \right) } \\ \mathrm{e}^{z} &= -1 - \mathrm{i} \\ \mathrm{e}^z &= \sqrt{2}\,\mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i} } \textrm{ where } n \in \mathbf{Z} \\ \mathrm{e}^{ x + \mathrm{i}\,y } &= \sqrt{2}\,\mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\, n \right) \,\mathrm{i} } \\ \mathrm{e}^x\,\mathrm{e}^{ \mathrm{i}\,y } &= \sqrt{2}\,\mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i} } \end{align*}$
so $\displaystyle \begin{align*} \mathrm{e}^{x} = \sqrt{2} \implies x = \ln{ \left( \sqrt{2} \right) } = \frac{1}{2}\ln{ \left( 2 \right) } \end{align*}$ and $\displaystyle \begin{align*} \mathrm{e}^{ \mathrm{i}\,y } = \mathrm{e}^{ \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \,\mathrm{i} } \implies y = -\frac{3\pi}{4} + 2\,\pi\,n \end{align*}$
Thus $\displaystyle \begin{align*} z = \frac{1}{2}\ln{ \left( 2 \right) } + \left( -\frac{3\,\pi}{4} + 2\,\pi\,n \right) \, \mathrm{i} \end{align*}$.
Is much as I hate it, yes that is what most applied mathematics textbooks teach as the correct answer to your question.
Only from my experience, they would expect $\displaystyle 0.5\ln(2)-135\bf{i}$ as the correct form.
@others, note that the question requires the $a+b\bf{i}$ form.
That implies that $a~\&~b$ are singular real numbers.
Really? Not $\displaystyle 0.5\ln(2)- (3\pi/4)\bf{i}$? I have never seen a text book, applied mathematics or otherwise, use degrees rather than radians here.
@others, note that the question requires the $a+b\bf{i}$ form.
That implies that $a~\&~b$ are singular real numbers.
I truly hope that is now the case. I have been out of the fight for ten years. But I hated be threaded with the lost of a large credit hour producing course they want engineers to use degrees.
Now that on I never gave in on that. Luckily it only seems to matter a small subset.