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Suppose $z=\frac{3+ai}{5+bi}$ for some constant $a$ and $b$. If $\operatorname{Re}(z)=3$ and $\operatorname{Im}(z)=4$, find $a$ and $b$.

I do understand that $z = x + iy$, and thus I have $z = 3 + 4i$.

Hence,

$$3 + 4i = \frac{3+ai}{5+bi}$$

May I find out how can I carry on from here?

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    $\begingroup$ The key is to convert the $(3+ia)/(5+ib)$ into the form $c+id$ and then identify it with $3+4i$. In order to do that, multiply $z$ by $(5-bi)/(5-bi)$. $\endgroup$
    – Atmos
    Commented Sep 3, 2021 at 8:59
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    $\begingroup$ Have you considered multiplying both the numerator and denominator of $z$ by the conjugate of $5+bi$? $\endgroup$
    – Ultimate Apple
    Commented Sep 3, 2021 at 9:00
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    $\begingroup$ Use fundamental properties of complex numbers $$\frac{a+bi}{c+di}=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i$$ (This comes from simple algebra) Then equalize coefficients and solve algebraic system of equation. $\endgroup$
    – lone student
    Commented Sep 3, 2021 at 9:02

2 Answers 2

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We know that $(3+4i)(5+bi)=3+ai$

Which reduces to $12-4b+(3b+20-a)i=0$.

Now, equating the real part and imaginary part of left hand side to $0$.

$12-4b=0$ hence, $b=3$.

Also, $3b+20-a=0$ gives $a=29$.

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    $\begingroup$ @user2661923 the OP has shown his/her efforts. He/she knows what Re(z) and Im(z) mean. He/she is just willing to know how to proceed after that equation. I am happy to help him/her. $\endgroup$
    – Ilovemath
    Commented Sep 3, 2021 at 9:36
  • $\begingroup$ I have no disagreement with you, which is why I upvoted to reverse the downvote. My comment was intended merely to explain why some other anonymous person downvoted your answer. $\endgroup$
    – user2661923
    Commented Sep 3, 2021 at 9:39
  • $\begingroup$ Hi @Ilovemath, thank you so much for the clear breakdown! I did manage to get to the equation of 12+3bi+20i-4b-ai = 0. But i didn't know how to proceed further. The fact that u extracted "i" as a multiple made things so clear for me. Thank you so much ! Sorry if i've broken the rule or seem like i did not put in any effort. But the fact is i've never passed math and now i'm being forced to pick up this in my university and its really stressful. Thank you so much! $\endgroup$
    – X-men
    Commented Sep 3, 2021 at 9:59
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whenever we encounter fractions with a complex number in the denominator, we multiply both parts of the fraction with the complex conjugate of the denominator to make the denominator a real number. (for $z=x+yi$, the conjugate is simply $\overline{z}=x-yi$). Multiplying with the conjugate both places will result in:

$\frac{(3+ai)(5-bi)}{(5+bi)(5-bi)}=\frac{(15+ab)+(5a-3b)i}{25-b^2}$

if you expand the fraction you will more easily see the difference in the real and imaginary part.

$\frac{15+ab}{25-b^2} + \frac{5a-3b}{25-b^2}i$

Then, as the excercise states, the real part is 3 and the imaginary part is 4. This is same same as saying

$\frac{15+ab}{25-b^2}= 3$ and $\frac{5a-3b}{25-b^2}=4$

From there, you can just solve those two equations for a and b yourself or use some CAS-tool.

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