Suppose I've the following:
$$\arg(2+i5-z) = 0.9$$
How can I find $z$? Wolfram says it's $-1.96$, but I didn't understand how to get this value.
EDIT: I wrote originally that the answer was 2, but it's in fact 0.90.
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1All the numbers on a ray from the origin have the save argument, so there isn't a unique value of $z$ that satisfies the equation. – saulspatz Nov 09 '20 at 20:48
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Wolfram says it's −1.96 Where is the corresponding link? – callculus42 Nov 09 '20 at 20:49
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Actually I copied wrong, it was supposed to be arg(2+i5-z) = 0.9. But here it's: https://www.wolframalpha.com/input/?i=arg%282%2Bi5+-+z%29+%3D+0.9 – FY Gamer Nov 09 '20 at 21:09
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Your question should be : How to find a Real number ... – hamam_Abdallah Nov 09 '20 at 21:24
1 Answers
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hint
Put $$z=a+ib$$
then
$$Z=2+5i-z=(2-a)+i(5-b)$$
with
$$\frac{5-b}{2-a}=\;\tan(0.9)$$
There are a lot of solutions.
If you look for real solution, with $ b=0$, you will get $$\frac{5}{2-a}=\tan(0.9)=1.26$$ So $$a=z=-1.96$$
hamam_Abdallah
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