Find the Value of z in z^{1+i}=4 using Logarithms

In summary, the conversation is about finding the value of z in the equation z^{1+i}=4. The participant suggests using the equation log(z1 z2)=log(z1)+log(z2) and re^{i\theta}=r(\cos\theta + i\sin\theta) to solve it. They present their attempt at a solution, but another participant points out a false assumption and suggests starting by taking logs.
  • #1
Terrell
317
26

Homework Statement


Find ##z## in ##z^{1+i}=4##. Is my solution correct

Homework Equations


##\log(z_1 z_2)=\log(z_1)+\log(z_2)## such that ##z_1, z_2\in \{z\in\Bbb{C} : (z=x+iy) \land (x\in\Bbb{R}) \land -\infty \lt y \lt +\infty\}##
##re^{i\theta}=r(\cos\theta + i\sin\theta)##

The Attempt at a Solution


Note ##4=4(\cos 0 +i\sin 0)##. Thus,
\begin{align}
r^{1+i}e^{i\theta(1+i)}&=r^{1+i}e^{-\theta}(\cos\theta + i\sin\theta)=4(\cos 0 +i\sin 0)=4\
r^{1+i}e^{-\theta}&=4 \quad \text{and}\quad \theta=0\\
\log_e(r^{1+i}e^{-\theta})&=\log_{e}(4)\\
\log_(r^{1+i})+\log(e^{-\theta})&=\log_{e}(4)\\
(1+i)\log_{e}(r)-\theta&=\log_{e}(4)\\
\log_{e}(r)&=\frac{\log_e(4)}{1+i}\\
\log_{e}(r)&=\log_{e}(4^{\frac{1}{1+i}})\\
r&=4^{\frac{1}{1+i}}
\end{align}
 
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  • #2
In step 1 you appear to be making the false assumption that r1+i is real. (Express it as elnr(1+i) and see.)
I think it would be easier if you start by taking logs
ln4 = (1+i)lnz
and see where that gets you.
 

Related to Find the Value of z in z^{1+i}=4 using Logarithms

What is the meaning of $z^{1+i} = 4$?

The equation $z^{1+i} = 4$ is asking for the complex number $z$ that, when raised to the power of $1+i$, results in a value of $4$. This can also be written as $(1+i)\log(z) = \log(4)$.

How do I solve $z^{1+i} = 4$?

To solve this equation, you can use logarithms to isolate the variable $z$. First, take the logarithm of both sides to get $\log(z) = \frac{\log(4)}{1+i}$. Then, use properties of logarithms to simplify the expression, giving $\log(z) = \frac{\log(4)}{1+i} = \frac{\log(4)}{\sqrt{2}} - i\frac{\log(4)}{\sqrt{2}}$. Finally, take the antilog of both sides to find the solution $z = e^{\frac{\log(4)}{\sqrt{2}} - i\frac{\log(4)}{\sqrt{2}}} = \frac{4}{\sqrt{2}}e^{-i\frac{\log(4)}{\sqrt{2}}}$.

What is the solution set for $z^{1+i} = 4$?

The solution set for $z^{1+i} = 4$ is a single complex number $z = \frac{4}{\sqrt{2}}e^{-i\frac{\log(4)}{\sqrt{2}}}$. This can also be written in polar form as $z = 4e^{-i\frac{\pi}{4}}$.

Can I solve $z^{1+i} = 4$ without using logarithms?

Yes, you can also solve this equation by rewriting it in polar form and using the definition of complex exponentiation. First, rewrite the equation as $z = 4^{1/(1+i)}$. Then, using the definition of complex exponentiation, we can write this as $z = 4^{1/(1+i)} = 4^{1/2}e^{i\theta}$, where $\theta = \tan^{-1}(1) = \frac{\pi}{4}$. Thus, the solution set is $z = 4^{1/2}e^{i\frac{\pi}{4}} = \frac{4}{\sqrt{2}}e^{i\frac{\pi}{4}}$.

How does the value of $i$ affect the solutions to $z^{1+i} = 4$?

The value of $i$ affects the solutions to $z^{1+i} = 4$ by changing the exponent of $z$. Specifically, the value of $i$ changes the argument of the complex number $z$ by adding a multiple of $i\pi/2$. This results in different solutions, as the value of $z$ will vary depending on the argument it is raised to. For example, if $i = 3$, then the solution set becomes $z = 4^{1/3}e^{i\frac{\pi}{12}}$, which is different from the original solution set of $z = 4^{1/2}e^{i\frac{\pi}{4}}$.

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