Complex no. (1 Viewer)

yanujw · Oct 23, 2021

Let
$z = a + ib$
It follows that $\overline{z} = a - ib$

If $z\overline{z} = 5$
then $a^2 + b^2 = 5$

If $\frac{z}{\overline{z}} = 0.2(3+4i)$
then $\frac{a^2 + 2abi - b^2}{a^2 + b^2} = 0.2(3+4i)$ (by multiplying using conjugates)

$=\frac{a^2 + 2abi - b^2}{5}$
So $a^2 + 2abi - b^2 = 3 + 4i$

Finally, simulatenously solve $a^2 - b^2 = 3$ and $2ab = 4$
I'll leave that final part as an exercise to you.

5uckerberg · Oct 23, 2021

=)(= said:
View attachment 32900

For this question, this is how I would see it z and its conjugate gives 5 automatically this should tell us that $x^{2}+y^{2}=5$ , 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 $3+4i$ and you know as a matter of fact there has to be something that gives us $3+4i$ .

Doing simultaneous equations
$x^{2}+y^{2}=5$
$x^{2}-y^{2}=3$
We see quickly that $2x^{2}=8, x=\pm{2}, y=\pm{1}$
Thus, $z=\pm({2+i})$

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 · Oct 23, 2021

yanujw said:
Let
$z = a + ib$
It follows that $\overline{z} = a - ib$

If $z\overline{z} = 5$
then $a^2 - b^2 = 5$

If $\frac{z}{\overline{z}} = 0.2(3+4i)$
then $\frac{a^2 + 2abi - b^2}{a^2 - b^2} = 0.2(3+4i)$ (by multiplying using conjugates)

$=\frac{a^2 + 2abi - b^2}{5}$
So $a^2 + 2abi - b^2 = 3 + 4i$

Finally, simulatenously solve $a^2 - b^2 = 3$ and $2ab = 4$
I'll leave that final part as an exercise to you.

I reckon this should be $a^{2}+b^{2}=5$ . One little mistake yet a vital one.

yanujw · Oct 23, 2021

5uckerberg said:
I reckon this should be $a^{2}+b^{2}=5$ . One little mistake yet a vital one.

That's true, its' edited now.
Interestingly, by making the same mistake twice I still got the correct solution lol.

CM_Tutor · Oct 23, 2021

@yanujw, you have stated that $a^2 - b^2 = 5$ and later that $a^2 - b^2 = 3$ , which has no possible solutions for $a$ and $b$ ...

Edit: Already noted and corrected, excellent

=)(= · Oct 23, 2021

I have another one slightly more difficult, part b please

vishnay · Oct 23, 2021

=)(= said:
I have another one slightly more difficult, part b please
View attachment 32903

$(z-\alpha)(z-\bar{\alpha})=z^2-2\Re(\alpha)z+\alpha\bar\alpha$

=)(= · Oct 23, 2021

how did you get the -2Re(a)z term ?

5uckerberg · Oct 23, 2021

=)(= said:
I have another one slightly more difficult, part b please
View attachment 32903

This question

=)(= said:
how did you get the -2Re(a)z term ?

$z(\alpha+\bar{\alpha})=2Re(\alpha)z$
Conjugates.

=)(= · Oct 23, 2021

5uckerberg said:
This question

$z(\alpha+\bar{\alpha})=2Re(\alpha)z$
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 · Oct 23, 2021

=)(= said:
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.

=)(= · Oct 23, 2021

ExtremelyBoredUser said:
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 · Oct 23, 2021

Note that the question is flawed.

The quadratic equation with roots at $\alpha$ and $\bar{\alpha}$ , where $\alpha$ is a complex number with a non-zeo imaginary part is

$(z - \alpha)(z - \bar{\alpha}) = z^2 - 2z\Re{(\alpha)} + \alpha\bar{\alpha} = z^2 - 2z\Re{(\alpha)} + |\alpha|^2 = 0$

but the question has the constant term as $\alpha|\bar{\alpha}|$ , which is a non-real complex number as $|\bar{\alpha}|$ is a real constant.

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

CM_Tutor · Oct 23, 2021

=)(= said:
ohh i get, is it because re(a) is just a since that is the real component of a?

$\text{Yes, if $\alpha = a + ib$, where $a,\ b \in \mathbb{R}$, then the real part of $\alpha$ is $\Re{(\alpha)} = \text{Re($\alpha$)} = a$ and the imaginary part of $\alpha$ is $\Im{(\alpha)} = \text{Im($\alpha$)} = b$.}$

=)(= · Oct 23, 2021

CM_Tutor said:
$\text{Yes, if $\alpha = a + ib$, where $a,\ b \in \mathbb{R}$, then the real part of $\alpha$ is $\Re{(\alpha)} = \text{Re($\alpha$)} = a$ and the imaginary part of $\alpha$ is $\Im{(\alpha)} = \text{Im($\alpha$)} = b$.}$

ohh makes sense i got mixed up between a and alpha

ExtremelyBoredUser · Oct 23, 2021

=)(= said:
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.

CM_Tutor · Oct 23, 2021

=)(= said:
ohh makes sense i got mixed up between a and alpha

Yes, I wondered if that might be the case, hence my clarification after I noticed the definition

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

=)(= · Oct 23, 2021

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?

ExtremelyBoredUser · Oct 23, 2021

=)(= said:
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?

They're the same thing. $|z|^2 = x^2 + y^2$

You can see;

$|z| = \sqrt(x^2 + y^2)$
$|z|^2 = x^2 + y^2$

where z is a complex number x + iy.

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