What is the modulo of z = e^ln(4) + i? Complex Numbers

[Say12]

z = e^ln(4) + i

= e^ln(4) * e^i
= 4 e^i

Can I use this form

e^i = cos(0) + isin(0)
= 1 + 0
= 1

So its 4?

 

[blamocur]

Can I use this form

I would start with proper use of parenthesis: is it e^ln(4)+i or e^(ln(4)+i) ?

 

[Dr.Peterson]

z = e^(ln(4) + i)

= e^ln(4) * e^i
= 4 e^i

Can I use this form

e^i = cos(0) + isin(0)
= 1 + 0
= 1

So its 4?

First, when you say "modulo" do you mean "modulus", also called "absolute value"? The term "modulo" is used in number theory for something very different.

Second, why did you use 0 in "e^i = cos(0) + i sin(0)". It may help if you write out the general formula before applying it.

 

[pka]

What is the modulo of z = e^ln(4) + i?​

Because [imath]e^{\log(4)}=4[/imath] your [imath]z=4+i[/imath].
Now if [imath]z=a+bi[/imath] then [imath]|z|=\sqrt{a^2+b^2}[/imath].
Can you complete?

 

[Steven G]

e^i = cos(0) + isin(0)
= 1 + 0
= 1

Since e⁰=1, it is not true that eⁱ=1

 

[Dr.Peterson]

Since e⁰=1, it is not true that eⁱ=1

Technically, that's not a good argument. Would you say that since e⁰ = 1, it can't be true that e²ⁱ = 1?

z = e^ln(4) + i

= e^ln(4) * e^i
= 4 e^i

I think it's reasonably clear that you meant, as I modified it when I answered, z = e^(ln(4) + i), since only that agrees with your next line.

But you need to respond to what we've said (and perhaps you have, but didn't get through moderation yet): What should you have used in place of 0 in this?

e^i = cos(0) + isin(0)
= 1 + 0
= 1

Just copy the formula you are applying, and you should see it.

 

[Steven G]

Technically, that's not a good argument. Would you say that since e⁰ = 1, it can't be true that e²ⁱ = 1?

I think you meant e^(2pi*i) = 1.
I knew as soon as I made that post that I was wrong.

 

[Dr.Peterson]

I think you meant e^(2pi*i) = 1.
I knew as soon as I made that post that I was wrong.

You passed the test.

I won't say whether it was an intentional test or not.