I need help to solve this problem:

Find two complex numbers that add up to 1+4i, their quotient is purely imaginary and the real part of one of them is -1

I've tried to start the exercise, but I honestly have no clue:

z + (1+4i) = -1 +ai

I also do not understand why in one of the solutions given by the lecturer is z = 2 + i (2- sqrt(2)) since the exercise says the real part must be -1.


Let $a,b,c,d\in\mathbb R$. You want to have

  • $(a+bi)+(c+di)=1+4i$;
  • $\frac{a+bi}{c+di}$ is purely imaginary;
  • $a=-1$ or $c=-1$.

From the first condition, you deduce that $c=1-a$ and that $d=4-b$. Therefore\begin{align}\frac{a+bi}{c+di}&=\frac{a+bi}{1-a+(4-b)i}\\&=\frac{a(1-a)+b(4-b)+\bigl(-a(4-b)+b(1-a)\bigr)i}{(1-a)^2+(4-b)^2}\end{align}Now, if $a=-1$, this last expression becomes$$\frac{-2+b(4-b)+(4+b)i}{(4-b)^2}.$$Since you want this number to be purely imaginary, you must take $b$ such that $b(4-b)=2$; take $b=2+\sqrt2$, for instance. So, one answer to your problem is: take the numbers $-1+(2+\sqrt2)i$ and $2+(2-\sqrt2)i$.

  • $\begingroup$ Thank you so much! I do not understand why you state that a = 1 or c = 1 $\endgroup$ – Evoked Oct 21 '17 at 11:39
  • $\begingroup$ @Evoked My mistake. I meant $-1$. $\endgroup$ – José Carlos Santos Oct 21 '17 at 11:40
  • $\begingroup$ Oh, know I see. Thank you for your help and your time! @José Carlos Santos $\endgroup$ – Evoked Oct 21 '17 at 12:02

$(a+bi)+(c+di)=1+4i\to a+c=1;\; c+d=4$

the quotient is imaginary

$$\frac{a+bi}{c+di}=\frac{a c+b d}{c^2+d^2}+\frac{b c-a d}{c^2+d^2}\,i$$

Real part is zero


and $a=-1$

so we have the system

$ \left\{ \begin{array}{l} a=-1 \\ a+c=1 \\ b+d=4 \\ a c+b d=0 \\ \end{array} \right. $

with two solutions

$(a = -1,b = 2-\sqrt{2},c = 2,d = 2+\sqrt{2}),(a = -1,b = 2+\sqrt{2},c = 2,d = 2-\sqrt{2})$



  • $\begingroup$ Ok, know I understand. Thank you so much @Raffaele ! $\endgroup$ – Evoked Oct 21 '17 at 12:03

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