Complex no. (1 Viewer)

yanujw

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Let

It follows that

If
then

If
then (by multiplying using conjugates)


So

Finally, simulatenously solve and
I'll leave that final part as an exercise to you.
 

5uckerberg

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For this question, this is how I would see it z and its conjugate gives 5 automatically this should tell us that , Now I suppose wither x=2, y=1 or x=1 and y=2. Here I see the fraction and oh look at this we need to rationalise the denominator and it is 5. After this, my attention goes towards and you know as a matter of fact there has to be something that gives us .

Doing simultaneous equations


We see quickly that
Thus,

Disclaimer. The fraction gave us a squared value so already you are doing a "find the square roots of a+ib," type of question.
 

5uckerberg

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Let

It follows that

If
then

If
then (by multiplying using conjugates)


So

Finally, simulatenously solve and
I'll leave that final part as an exercise to you.
I reckon this should be . One little mistake yet a vital one.
 

CM_Tutor

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@yanujw, you have stated that and later that , which has no possible solutions for and ...

Edit: Already noted and corrected, excellent :)
 

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I have another one slightly more difficult, part b please
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how did you get the -2Re(a)z term ?
 

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This question


Conjugates.
ohh i get it because the conjugates are identical to the original, so its just z(2a), would you have to put Re(a) since it said the coefficents were real in the q?
 

ExtremelyBoredUser

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ohh i get it because the conjugates are identical to the original, so its just z(2a), would you have to put Re(a) since it said the coefficents were real in the q?
I mean you could do it algebraically to convince yourself.

let a be a complex number, a+ib. a conjugate is simply a - ib.

a + ib + a - ib = 2a

and 2a is simply double the real component of a complex number, a + ib "2Re(a)". It would be considered a coefficient since you would have factored it from z.
 

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I mean you could do it algebraically to convince yourself.

let a be a complex number, a+ib. a conjugate is simply a - ib.

a + ib + a - ib = 2a

and 2a is simply double the real component of a complex number, a + ib "2Re(a)". It would be considered a coefficient since you would have factored it from z.
ohh i get, is it because re(a) is just a since that is the real component of a?
 

CM_Tutor

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Note that the question is flawed.

The quadratic equation with roots at and , where is a complex number with a non-zeo imaginary part is


but the question has the constant term as , which is a non-real complex number as is a real constant.

I am sure the question is seeking one of the above equations but, as written, it is impossible.
 

ExtremelyBoredUser

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ohh makes sense i got mixed up between a and alpha
if thats the case then try using x+iy or any other variables that are more distinguishable so you don't subconsciously interpret one variable as another unless specified.
 

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i was also wondering why does z multiplied by its conjugate give the absolute value of z ^2, shouldn't it just be x^2 + y^2?
 

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