April 30th, 2017, 03:04 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  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, 05:55 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, 08:55 AM  #3 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 16th, 2017, 08:35 PM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,185 Thanks: 90 
$\displaystyle a^{z}=e^{zlna}\\ 2^{z}=e^{xln2}=1\\ x=0$ (A) 
May 16th, 2017, 10:14 PM  #5 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 17th, 2017, 12:26 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,185 Thanks: 90  
May 17th, 2017, 12:45 AM  #7 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 17th, 2017, 07:43 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,185 Thanks: 90 
$\displaystyle z=re^{i\theta}$, z=r 
May 17th, 2017, 08:46 PM  #9 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 18th, 2017, 12:38 PM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,185 Thanks: 90  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  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Basis for Complex Vector Space over the Complex and then the Real Fields  whh2  Linear Algebra  1  April 30th, 2015 05:00 AM 
Can complex numbers (and properties of complex numbers) be..  jonas  Complex Analysis  2  October 13th, 2014 04:03 PM 
Complex complex numbers  Tutu  Algebra  11  June 26th, 2012 02:36 PM 
Complex Solutions to A "Complex" quadratic  Ajihood  Complex Analysis  1  March 16th, 2009 10:02 AM 
Complex  sunsetsun  Complex Analysis  0  June 9th, 2008 01:02 PM 