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 April 30th, 2017, 02:04 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4 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
 April 30th, 2017, 04:55 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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.
May 16th, 2017, 07:55 AM   #3
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 Originally Posted by agentredlum 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 ??

 May 16th, 2017, 07:35 PM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle a^{z}=e^{zlna}\\ |2^{z}|=e^{xln2}=1\\ x=0$ (A)
May 16th, 2017, 09:14 PM   #5
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 Originally Posted by zylo $\displaystyle a^{z}=e^{zlna}\\ |2^{z}|=e^{xln2}=1\\ x=0$ (A)
I didn't understand it well. Can you please elaborate it for me.
Thank you

May 16th, 2017, 11:26 PM   #6
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 Originally Posted by zylo $\displaystyle a^{z}=e^{zlna}\\ |2^{z}|=e^{xln2}=1\\ x=0$ (A)
Definition of $\displaystyle a^{z}$

If $\displaystyle e^{z}=e^{x}e^{iy},\\ |e^{z}|=e^{x}$
Re(z)=x

May 16th, 2017, 11:45 PM   #7
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 Originally Posted by zylo 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 $\displaystyle e^{x}$ ?

 May 17th, 2017, 06:43 AM #8 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle z=re^{i\theta}$, |z|=r
May 17th, 2017, 07:46 PM   #9
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 Originally Posted by zylo $\displaystyle z=re^{i\theta}$, |z|=r
I have a doubt... does $|a^z|$ implies that $|a^{|z|}|$ ?

May 18th, 2017, 11:38 AM   #10
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 Originally Posted by Lalitha183 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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