I am trying to solve this equation: $3z - i\overline{z}=7-5i$ and I am stuck.
I start with changing $z=a+bi$ and $\overline{z}=a-bi$
$3(a+bi) - i(a-bi)=7-5i$
$3a+3bi - (ia-bi^2)=7-5i$
$3a+3bi - ia+bi^2=7-5i$
$3a+3bi - ia-b=7-5i$
And now I don't really know what to do next but I have tried this:
$3a- ia+3bi -b=7-5i$
$a(3- i)+b(3i -1)=7-5i$
$a(3- i)+b(-1+3i )=7-5i$
How do I solve this? Thanks.
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Ridertvis
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4Compare real and imaginary parts to get two simultaneous equations in $a,b$. – preferred_anon Jan 27 '23 at 15:53
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4Does this answer your question? Complex numbers: With conjugate – Anne Bauval Jan 27 '23 at 15:55
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2Although slightly different, you may try $b=\frac{7-5 i}{3i-1},a=0$ – Тyma Gaidash Jan 27 '23 at 16:01
2 Answers
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No need to go with the real and imaginary parts. If you know that $$ 3z-i\bar{z}=7-5i $$ then you know that the same relation holds also for the conjugates $$ 3\bar{z}+iz=7+5i $$ Multiply the second equation by $i$ and the first equation by $3$: \begin{cases} 9z-3i\bar{z}=21-15i \\[6px] -z+3i\bar{z}=7i-5 \end{cases} Can you finish?
egreg
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1(+1) but this solution was present in the duplicate previously mentionned above. – Anne Bauval Jan 27 '23 at 16:38
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1@AnneBauval I didn't see it, but I believe that elimination is better (and faster). – egreg Jan 27 '23 at 16:52
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Collect the real terms and imaginary terms like so:
$$(3a-b)-(a+b)i = 7-5i$$
Now, compare the real terms on the left hand side to real terms on the right hand side: $3a-b=7$
Now, do the same for the imaginary terms: $a+b=5$
You can now solve these simultaneously to get the values of $a$ and $b$.
Alain Remillard
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user743392
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This solution is wrong and anyway useless: better look at the duplicate previously mentionned above. – Anne Bauval Jan 27 '23 at 16:33
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@AnneBauval Why is it wrong?I got the right answer when I used the method provided in this answer. – Ridertvis Jan 27 '23 at 17:33
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1The right answer is $a=2,b=-1,$ so $a+b=1\ne5.$ Look at the duplicate linked above. – Anne Bauval Jan 27 '23 at 17:37
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