
Complex Analysis Complex Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
December 10th, 2016, 01:23 AM  #1 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 46 Thanks: 5  simple looking complex equation  but I think it is not simple
$\displaystyle \leftz\right+z=2+i$ I tried with the Euler's form and got: $\displaystyle \leftz\right\cdot e^0+\leftz\right\cdot e^{i\:\phi }=\sqrt{5}\cdot e^{\frac{\pi }{6}}$ $\displaystyle \leftz\right\left(e^0+e^{i\:\phi \:}\right)=\sqrt{5}\cdot e^{\frac{\pi }{6}}$ $\displaystyle \leftz\right=\sqrt{5}$ $\displaystyle e^0+e^{i\cdot \phi }=e^{i\cdot \frac{\pi }{6}}$ I do not know how to solve this exponential equation. The solution of this task is: 3/4+i... I see also that I calculated the absolute value wrongly. Can you help me with any useful ideas and corrections please? Last edited by skipjack; December 17th, 2016 at 02:59 AM. 
December 10th, 2016, 01:31 AM  #2 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 46 Thanks: 5 
Well, I managed to solve it. It is quite easy if you start solving it correctly. $\displaystyle \leftz\right2=iz$ $\displaystyle \leftz\right2=a+i\left(1b\right)$ $\displaystyle b=1$ $\displaystyle \sqrt{a^2+1^2}2=a$ $\displaystyle \sqrt{a^2+1^2}=2a\:\:\backslash \cdot \left(\right)^2$ $\displaystyle a=\frac{3}{4}$ $\displaystyle z=\frac{3}{4}+i$ Last edited by skipjack; December 17th, 2016 at 03:00 AM. 
December 10th, 2016, 01:32 AM  #3 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 46 Thanks: 5 
But I would still like to know where I have made a mistake in my first attempt. Thanks!

December 16th, 2016, 08:53 PM  #4 
Member Joined: Dec 2016 From:  Posts: 36 Thanks: 7 
Well, for the exponential you don't know $\phi$, which is the phase associated to z. With your first method, you can only calculate the modulus of z (incorrectly), but there are infinite complex numbers with that same value of $z$.
Last edited by skipjack; December 17th, 2016 at 03:00 AM. 
December 16th, 2016, 10:54 PM  #5  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,166 Thanks: 173 
You used the substitution z = a + bi Quote:
Nice one. Last edited by skipjack; December 17th, 2016 at 03:01 AM.  

Tags 
complex, equation, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
kind of silly/simple question with probably a simple answer.  GumDrop  Math  4  October 4th, 2016 04:34 PM 
Simple geometry problem with a complex context  Brightstar  Computer Science  6  February 9th, 2012 04:14 PM 
Simple complex conjugate problem  dunn  Complex Analysis  3  January 16th, 2012 03:19 AM 
Simple, not so simple question about areas of triangles  jkh1919  Algebra  1  November 20th, 2011 09:14 AM 
Simple Equation I need help with?  cvspeed  Algebra  1  December 4th, 2008 05:43 PM 