Find two complex numbers such that their sum is $\ 1+4i$ , their quotient is purely imaginary, and the real part of one of them is$\ -1 $.

I have tried to create two complex values called: $\ z=-1+ci $ and $\ w=a+bi $ and computing its sum knowing the result. Therefore: $\ -1+ci+a+bi=1+4i \rightarrow a=2 \land c+b=4 $.

Then, we know that $\ \frac{-1+ci}{2+bi}=xi $ being$\ x $ the variable that I included to form every purely imaginary complex number. So, here is when i got stuck.

Thank you in advance!

As has being answered, you just need to use the relation $\ c=4-b \land z=xiw $. Solving for b will output a value of $\ b=2+\sqrt2 \lor b=2-\sqrt2 $ resulting in $\ c=2-\sqrt2 $ if $\ b=2+\sqrt2 $ and $\ c=2+\sqrt2 $ if $\ b=2-\sqrt2 $.

Its just a matter of patience (so don't be like me, be patient).

  • $\begingroup$ if their quotient is purely imaginary then $w = ki z$ $\endgroup$ – Doug M Jun 6 '18 at 16:48

Because $b+c=4$, we may as well write $c=4-b$. Hence you have the purely imaginary $$\frac{-1+(4-b)i}{2+bi}=\frac{-1+(4-b)i}{2+bi}\frac{2-bi}{2-bi}=\cdots$$

  • $\begingroup$ That's where i lose myself. Tried continuing and somehow arrived to: $\ \frac{-2+4b+b^2+(8+3b)i}{4+b} $ having a result of $ b= -2+-\sqrt(6)$ and its respective c. But when i try the division, the result is not purely imaginary. $\endgroup$ – ManuVS Jun 6 '18 at 17:11
  • $\begingroup$ 1. The denominator is $4+b^2$. 2. Forget about the denominator, for this to be purely imaginary you need $-2-4b+b^2=0$. $\endgroup$ – vadim123 Jun 6 '18 at 17:32
  • $\begingroup$ @vadim123 This is a minor point. As you stated, for the numerator to be purely imaginary means the real part needs to be $0$. However, the equation for that would be $-2 + 4b - b^2 = 0$ since $i(-bi) = b$. $\endgroup$ – John Omielan Jun 9 at 1:00

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