Solving for Log(z) and log(z): -9+2i

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In summary, The values of log(z) and Log(z) for z=-9+2i can be determined using the multivalued logarithm function, with the principal branch of the function being Log(z)= Log|z| + i Arg(z). For z=-9+2i, the values are \log (-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right) + 2k\pi i for all k\in\mathbb{Z} and \mbox{Log}(-9+2i) = \mbox{Log} \sqrt{85} + i
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suspenc3
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Hi, I'm kinda confused with the following, I must be looking at it wrong

Determinie all values of log(z) and Log(z):

a.) -9+2i

im confused when it comes to the angle. It is going to be in the form [tex]re^{i \theta}[/tex]. where [tex]r= \sqrt{85}[/tex]

Thanks.
 
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  • #2
Know that [tex]\log (z)= \mbox{Log} (z) +2k\pi i[/tex] for [tex]k\in\mathbb{Z}[/tex] is the multivalued logarithm function, and that [tex]\mbox{Log} (z)= \mbox{Log} |z| i\Arg (z)+2k\pi i[/tex] is the principal branch of the multivalued function [tex]\log (z)[/tex].

Now if [tex]z=re^{i\theta}[/tex] then [tex]\log (z) =\log (re^{i\theta}= \mbox{Log} (r) + i\theta + 2k\pi i[/tex] and hence for [tex]z=-9+2i[/tex] we have [tex]r=\sqrt{85}[/tex] and [tex]\theta = \tan ^{-1} \left( -\frac{9}{2}\right)[/tex] so that

[tex]\log (-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right) + 2k\pi i, \mbox{ for } k\in\mathbb{Z}[/tex]

is all values of [tex]\log (-9+2i)[/tex] and [tex]\mbox{Log} (-9+2i)[/tex] is the principal value of [tex]\log (-9+2i)[/tex] (so put [tex]k=0[/tex] in the above to get

[tex]\mbox{Log}(-9+2i) = \mbox{Log} \sqrt{85} + i \tan ^{-1} \left( -\frac{9}{2}\right)[/tex].
 
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Related to Solving for Log(z) and log(z): -9+2i

What is the difference between solving for log(z) and log(z)?

Solving for log(z) and log(z) both involve finding the logarithm of a complex number z. However, log(z) refers to the natural logarithm, while log(z) can refer to any logarithm with a specified base. The steps for solving these equations may differ depending on the given base.

How do I solve for log(z) and log(z) using algebraic methods?

To solve for log(z) and log(z) algebraically, you can use the properties of logarithms, such as the power rule and the product rule. First, rewrite the complex number z in exponential form, then apply the appropriate logarithmic rule to isolate the variable. Finally, solve the resulting equation using basic algebra.

Can I use a calculator to solve for log(z) and log(z)?

Yes, you can use a calculator to solve for log(z) and log(z). Make sure to use the appropriate function for the given base, such as the "ln" button for the natural logarithm. Keep in mind that some calculators may not accept complex numbers as input, so you may need to convert the number to polar form before using the calculator.

Are there any restrictions when solving for log(z) and log(z)?

Yes, there are restrictions when solving for log(z) and log(z). The logarithm of a non-positive number is undefined, so the complex number z must have a positive real component. Additionally, if solving for log(z) with a base other than e, the complex number z cannot have a complex component.

Can I use the properties of logarithms to simplify the expression log(zazb)?

Yes, you can use the product rule of logarithms to simplify the expression log(zazb) to log(za+b). However, if the base of the logarithm is not specified, you may need to use the power rule as well.

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