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 December 10th, 2016, 12:23 AM #1 Member   Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 simple looking complex equation - but I think it is not simple $\displaystyle \left|z\right|+z=2+i$ I tried with the Euler's form and got: $\displaystyle \left|z\right|\cdot e^0+\left|z\right|\cdot e^{i\:\phi }=\sqrt{5}\cdot e^{\frac{\pi }{6}}$ $\displaystyle \left|z\right|\left(e^0+e^{i\:\phi \:}\right)=\sqrt{5}\cdot e^{\frac{\pi }{6}}$ $\displaystyle \left|z\right|=\sqrt{5}$ $\displaystyle e^0+e^{i\cdot \phi }=e^{i\cdot \frac{\pi }{6}}$ 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 01:59 AM. December 10th, 2016, 12:31 AM #2 Member   Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 Well, I managed to solve it. It is quite easy if you start solving it correctly. $\displaystyle \left|z\right|-2=i-z$ $\displaystyle \left|z\right|-2=-a+i\left(1-b\right)$ $\displaystyle b=1$ $\displaystyle \sqrt{a^2+1^2}-2=-a$ $\displaystyle \sqrt{a^2+1^2}=2-a\:\:\backslash \cdot \left(\right)^2$ $\displaystyle a=\frac{3}{4}$ $\displaystyle z=\frac{3}{4}+i$ Thanks from agentredlum Last edited by skipjack; December 17th, 2016 at 02:00 AM. December 10th, 2016, 12:32 AM #3 Member   Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 But I would still like to know where I have made a mistake in my first attempt. Thanks! December 16th, 2016, 07:53 PM #4 Member   Joined: Dec 2016 From: - Posts: 62 Thanks: 10 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 02:00 AM. December 16th, 2016, 09:54 PM   #5
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You used the substitution z = a + bi

Quote:
 Originally Posted by srecko Well, I managed to solve it. It is quite easy if you start solving it correctly. $\displaystyle \left|z\right|-2=i-z$ $\displaystyle \left|z\right|-2=-a+i\left(1-b\right)$ $\displaystyle b=1$ ...
I agree, the expression on the left is a real number therefore b must be 1. After that, it is simple.

Nice one. Last edited by skipjack; December 17th, 2016 at 02:01 AM. Tags complex, equation, simple 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 GumDrop Math 4 October 4th, 2016 04:34 PM Brightstar Computer Science 6 February 9th, 2012 03:14 PM dunn Complex Analysis 3 January 16th, 2012 02:19 AM jkh1919 Algebra 1 November 20th, 2011 08:14 AM cvspeed Algebra 1 December 4th, 2008 04:43 PM

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