Consider the equation [math]z^2+az+b=0[/math]in the set of complex numbers, where a and b are real numbers. How many pairs (a, b) exist such that the equation has two solutions z1 and z2 satisfying [math]z_1-3=(1-|z_2|)i[/math]I tried Isolating [math]z_1=3+(1-|z_2|)i[/math]Then I think z1 and z2 are complex solutions so z2 will be [math]z_2=3-(1-|z_2|)i[/math]I used Viète [math]z_1z_2=b=9+(1-|z_2|)^2[/math][math]z_1+z_2=-a=3<=>a=-3[/math]I then set [math]z_1=a+bi[/math][math]z_2=c+di[/math]so [math]b=10-2\sqrt{c^2+d^2}+c^2+d^2[/math]Subtitute[math]z_1,z_2[/math] into [math]z_1-3=(1-|z_2|)i[/math][math]<=>a^2+b^2=10-2\sqrt{c^2+d^2}+c^2+d^2<=>a^2+b^2=b<=>b^2-b+9=0,(a=-3)[/math]I solved this and it has no real solutions so after all I messed up the whole process,Please help me^^
Solve for real numbers a and b: How many pairs solve z^2 + az + b = 0 in the complexes?
- Thread starter Violette
- Start date
Edits : I tried again and used [math]z_1=a+bi[/math][math]z_2=a-bi[/math]subtitute into [math]z_1-3=(1-|z_2|)i[/math]We get [math](a-3)+bi=(1-\sqrt{a^2+b^2})i<=>[/math][math]a-3=0<=>a=3[/math][math]b=1-\sqrt{a^2+b^2}<=>\sqrt{a^2+b^2}=1-b<=>\sqrt{3^2+b^2}=1-b[/math]Then we get [math]9+b^2=1-2b+b^2<=>b=-4[/math]but there's only one (a;b) the answer to the question is 3 where did I mistake?
oh my bad I didn't mean to use the same a,b of the equation i mean z=x+yi (x,y are real numbers) the complex form and I think the 5th equation is not wrong i^2=-1
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,104
You are right -- my bad
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,104
But how about [imath]z_1+z_2[/imath] ?You are right -- my bad
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,619
I agree with your work; untangling things, you have found that [imath]z_1=3-4i[/imath] so [imath]z_2=3+4i[/imath], and we can work backward to find that the equation is [imath]z^2-6z+25=0[/imath]. This gives only one pair [imath](a,b)=(-6,25)[/imath]. This is the result of your work, when fixed up, correct?
Now, you say you were told the answer is 3, meaning there should be 3 ordered pairs (a,b) that work, right?
Could it be that that answer is wrong?
(I'm not yet claiming that my answer can't be wrong!)
yes (-6;25) is one of the pairs.After a few tries I realized that was just case 1 when [math]delta=b^2-4ac<0[/math](the formula of delta), we will get 2 complex roots.I missed case 2 when [math]delta>=0[/math]which means we will get [math]a^2-4b>=0[/math] the condition for pair (a;b)
This means [math]z_1;z_2[/math] are real number[math]=> z_1-3=0[/math][math]1-|z_2|=0[/math]We get 2 pairs of [math](z_1;z_2)=(3;1);(3;-1)[/math]using viette for 2 pairs of [math](z_1;z_2)[/math]we get [math](a;b)=(3;-4);(-3;-2)[/math]both pairs satisfy the condition.In conclusion we will have 3 pairs of (a;b) which is the answer to the problem^^
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,619
Ah! In other words, what we were doing assumed that the two zeros are non-real, and therefore are complex conjugates. If we assume instead that they are real, then the equation implies that [imath]|z_2|=1[/imath], so [imath]z_2=\pm1[/imath], and taking each choice yields a solution.
There's a lesson to be learned there ...