# Thread: Complex numbers

2. ## Re: logarithmic functions and their graphs

Originally Posted by retro
Although I belong to a group of analysts who object to using the radical signs with complex numbers here is some help.
$\large{(-3\mathcal{i})(3\mathcal{i})=9}$

$\large{(2+\mathcal{i})(2-\mathcal{i})=5}$

3. ## Re: logarithmic functions and their graphs

How would u represent it without using radical signs

4. ## Re: logarithmic functions and their graphs

Originally Posted by retro
How would u represent it without using radical signs
Use the 1/2 power: $\displaystyle \left(\frac{(-3i)(3i)}{(2+ i)(2- i)}\right)^{1/2}$

What I know: common logarithms use base-10.

What I think I know: a logarithm is an exponent.

I don't know how to parlay my limited knowledge into a solution, please help
What was wrong with doing exactly what you were told to do?
"Choose two numbers". Okay, I choose 3 an 7. "Take their common logarithms" log(3)= 0.4771 and log(7)= 0.8451. "Add the logarithms" log(3)+ log(7)= 0.4771+ 0.8451= 1.3222. "Take the logarithm of their product". 3*7= 21 and log(21)= 1.3222. Hmmm! Now choose three numbers and try it again!

In the future post two different questions in different threads. People who have already responded to the first question might not look at the thread again and not see the new question.

5. ## Re: logarithmic functions and their graphs

Originally Posted by retro
How would u represent it without using radical signs
If you are still asking about $\sqrt{\dfrac{(-3i)(3i)}{(2+i)(2-i)}}$ then

Originally Posted by Plato
Although I belong to a group of analysts who object to using the radical signs with complex numbers here is some help.
$\large{(-3\mathcal{i})(3\mathcal{i})=9}$
$\large{(2+\mathcal{i})(2-\mathcal{i})=5}$
So that we have
\begin{align*}\sqrt{\dfrac{(-3i)(3i)}{(2+i)(2-i)}}&=\sqrt{\dfrac{9}{5}} \\&=\dfrac{3}{\sqrt 5}\\&=\dfrac{3\sqrt 5}{5} \end{align*}