How do i find the principal form of this complex number

iampaul

z= (10^-j+e^logj)²


I tried expanding but it just made it more complicated.
help please!!

iampaul

the principal form is defined in my book as
ln(z)=ln(r)+jθ

I like Serena

Hi iampaul!

To find the principal form you need to find the length and the angle of z.
First step is to find the length and angle of 10^-j.
And also of e^logj.
Do you know how?

iampaul

How?
That's exactly what my problem is.
10^-j and e^logj
aren't similar to any of the forms discussed in school.
Can you give another hint?

I like Serena

Let's try and find ln(z)=ln(r)+jθ for 10^-j and for e^logj.

ln(10^-j) = -j ln(10) = ln(1) - j ln(10)

So what is the length r and what is the angle θ for 10^-j?


Let's break e^logj apart a little further.

First we look at "j".
Did you know that "j" has length 1 and angle pi/2?
In other words, we have:
$$j = e^{j {\pi \over 2}}$$

Can you take the log of j?

And if so, find the length and angle of e^logj?

iampaul

thanks!!