April 30th, 2017, 02:04 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 159 Thanks: 1  Complex Value
Hello All I request you to correct me if the following answer is wrong $2^z = 1$ for a nonzero 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: 3,337 Thanks: 223 
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 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 159 Thanks: 1  
May 16th, 2017, 07:35 PM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 
$\displaystyle a^{z}=e^{zlna}\\ 2^{z}=e^{xln2}=1\\ x=0$ (A) 
May 16th, 2017, 09:14 PM  #5 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 159 Thanks: 1  
May 16th, 2017, 11:26 PM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87  
May 16th, 2017, 11:45 PM  #7 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 159 Thanks: 1  
May 17th, 2017, 06:43 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 
$\displaystyle z=re^{i\theta}$, z=r 
May 17th, 2017, 07:46 PM  #9 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 159 Thanks: 1  
May 18th, 2017, 11:38 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87  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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