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April 30th, 2017, 02:04 AM   #1
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Complex Value

Hello All
I request you to correct me if the following answer is wrong
$|2^z| = 1$ for a non-zero complex number z then which one of the following is necessarily true.

(A) $Re(z)$ = 0. (B) $|z|$ = 1. (C) $Re(z)$ = 1. (D) No such $z $ exists.

I am thinking like if the complex value of $ z$ is $1+i^2$ then

1+(-1)=0 ( since $i^2 = -1$)

$2^0=1$ is satisfied.
So I think the answer would be option (C).

Kindly check and let me know if I'm wrong.

Thank you
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April 30th, 2017, 04:55 AM   #2
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According to your thinking ...

What if $ \ \ \ z = 2 + 2i^2 \ \ \ $ ?

Then $ \ \ \ z = 0 \ \ \ $ also but $ \ \ \ Re(z) \neq 1 $

Also , $ \ \ \ z = a + bi \ \ \ $ NOT $ \ \ \ a + bi^2$

Think about it.

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May 16th, 2017, 07:55 AM   #3
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Quote:
Originally Posted by agentredlum View Post
According to your thinking ...

What if $ \ \ \ z = 2 + 2i^2 \ \ \ $ ?

Then $ \ \ \ z = 0 \ \ \ $ also but $ \ \ \ Re(z) \neq 1 $

Also , $ \ \ \ z = a + bi \ \ \ $ NOT $ \ \ \ a + bi^2$

Think about it.

Does it mean there is no such $z$ exists ??
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May 16th, 2017, 07:35 PM   #4
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$\displaystyle a^{z}=e^{zlna}\\
|2^{z}|=e^{xln2}=1\\
x=0$

(A)
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May 16th, 2017, 09:14 PM   #5
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Quote:
Originally Posted by zylo View Post
$\displaystyle a^{z}=e^{zlna}\\
|2^{z}|=e^{xln2}=1\\
x=0$

(A)
I didn't understand it well. Can you please elaborate it for me.
Thank you
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May 16th, 2017, 11:26 PM   #6
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Quote:
Originally Posted by zylo View Post
$\displaystyle a^{z}=e^{zlna}\\
|2^{z}|=e^{xln2}=1\\
x=0$

(A)
Definition of $\displaystyle a^{z}$

If $\displaystyle e^{z}=e^{x}e^{iy},\\
|e^{z}|=e^{x} $
Re(z)=x
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May 16th, 2017, 11:45 PM   #7
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Quote:
Originally Posted by zylo View Post
Definition of $\displaystyle a^{z}$

If $\displaystyle e^{z}=e^{x}e^{iy},\\
|e^{z}|=e^{x} $
Re(z)=x
Is $\displaystyle |e^{z}| $ equals to $\displaystyle e^{x} $ ?
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May 17th, 2017, 06:43 AM   #8
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$\displaystyle z=re^{i\theta}$,
|z|=r
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May 17th, 2017, 07:46 PM   #9
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Quote:
Originally Posted by zylo View Post
$\displaystyle z=re^{i\theta}$,
|z|=r
I have a doubt... does $|a^z|$ implies that $|a^{|z|}|$ ?
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May 18th, 2017, 11:38 AM   #10
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Quote:
Originally Posted by Lalitha183 View Post
I have a doubt... does $|a^z|$ implies that $|a^{|z|}|$ ?
No

$\displaystyle a^{z}= e^{z\ln a}$, This is a definition. The minute you start to think about it you are cooked. It is defined this way because e$\displaystyle ^{w}$ is defined but a$\displaystyle ^{w}$ is not.


$\displaystyle a^{|z|}= e^{|z|\ln a}$ is a real number (|z| is real}.
$\displaystyle a^{z}= e^{z\ln a} = e^{(x+iy)\ln a}$ is a complex number.
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