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December 10th, 2016, 12:23 AM   #1
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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.
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December 10th, 2016, 12:31 AM   #2
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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.
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December 10th, 2016, 12:32 AM   #3
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But I would still like to know where I have made a mistake in my first attempt. Thanks!
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December 16th, 2016, 07:53 PM   #4
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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.
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December 16th, 2016, 09:54 PM   #5
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You used the substitution z = a + bi

Quote:
Originally Posted by srecko View Post
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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