Find two different solutions to the equation a + b i = 4 + 5 i
Hello gayabee. First of all Welcome To OpenStudy.
Why thank you :)
i m wondering about the second solution
Yeah, I got the first ones where a= 4 and b = 5
I have : \(a + bi = 4 + 5 i \) By equating real parts of the both sides : Real Part of (a+bi) = Real part of (4+ 5i) Real part of (a+bi) = a and Real part of (4+5i) = 4 so a = 4
yes and by equating imaginary parts : b = 5
But apparently there are 2 sets of solutions that being the first set.
There will only be a single set of soln .
i also think that
I thought that too but apparently not. I'm on my uni site doing an online quiz
Do you have options?
where a and b are real or complex numbers, and neither of a or b is zero. 1 and 2. Solution One: a = b = 3 and 4. Solution Two: a = b =
That's all I've got
Hmm, put in the first one , a = 4 and b = 5 and let the second one be empty. is it allowed to do so?
Can you snap the question and attach here?
we could extend this to another space of complex numbers. Do you understand Re and Im graphs?
yes i do
@primeralph are you sure that you can arrive to another different set of solution for a and b in a + b = 4 + 5i ? [ we have one soln as : a = 4 and b = 5 ]
@gayabee please snap the quest. and post here.
i can't seem to snap it but this is exactly what it is ----------------------------------------------------------------- Find two different solutions to the equation a + b i = 4 + 5 i where a and b are real or complex numbers, and neither of a or b is zero. 1 and 2. Solution One: a = b = 3 and 4. Solution Two: a = b =
@mathslover simply speaking, no. Analytically speaking, yes. @gayabee perfect, but I must warn you, this isn't really an answer unless it is specially requested.
all right. got that. thank you! @primeralph
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