My Math Forum simple looking complex equation - but I think it is not simple

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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.

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