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 Complex Analysis Complex Analysis Math Forum

 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. Tags complex Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post whh2 Linear Algebra 1 April 30th, 2015 04:00 AM jonas Complex Analysis 2 October 13th, 2014 03:03 PM Tutu Algebra 11 June 26th, 2012 01:36 PM Ajihood Complex Analysis 1 March 16th, 2009 09:02 AM sunsetsun Complex Analysis 0 June 9th, 2008 12:02 PM

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