Complex question (1 Viewer)

synthesisFR · Jul 31, 2023

their method

synthesisFR · Jul 31, 2023

like they say "this formula should not be memorised" and proceed to do those examples using that formula
da hell bruv

SadCeliac · Jul 31, 2023

synthesisFR said:
like they say "this formula should not be memorised" and proceed to do those examples using that formula
da hell bruv

"it's intuitive"

synthesisFR · Jul 31, 2023

can someone do those 2 examples normally so i know what to do
when you can do q16's but not solve a quadratic equation bc ur kinda messed up af like me

SadCeliac · Jul 31, 2023

from skimming it (sorry if I'm repeating what someone else said) they use the discriminant? I have no clue I need to work this out on paper

synthesisFR · Jul 31, 2023

Gabriel1210 said:
use the discriminant?

U CANT JUST SAY THAT

JUST CURE CANCER OKAY

SadCeliac · Jul 31, 2023

synthesisFR said:
U CANT JUST SAY THAT

JUST GET CANCER OKAY

BRO IM NOT WALTER WHITE CHILL

synthesisFR · Jul 31, 2023

anyways im gonna go to sleep
@carrotsss ill leave this with you then

SadCeliac · Jul 31, 2023

synthesisFR said:
anyways im gonna go to sleep
@carrotsss ill leave this with you then

^

carrotsss · Jul 31, 2023

synthesisFR said:
can someone do those 2 examples normally so i know what to do
when you can do q16's but not solve a quadratic equation bc ur kinda messed up af like me

Basically you just solve it like a normal quadratic, except what is in the root is a complex number, so you will need to find the root of the complex number to determine it. To determine this root, you can either complete the square (extremely elegant and fun, but unnecessary) or find a result by equating real and imaginary parts in (x+iy)^2=discriminant. Then you equate real and imaginary parts, and from there it’s just algebra to find x and y, after which you can substitute in (x+iy)^2, which can be rooted to finally determine the value of z. You can also convert the discriminant to exponential form to make the root easier but that’s not always an option if it’s not a round argument obviously

synthesisFR · Jul 31, 2023

carrotsss said:
Basically you just solve it like a normal quadratic, except what is in the root is a complex number, so you will need to find the root of the complex number to determine it. To determine this root, you can either complete the square (extremely elegant and fun, but unnecessary) or find a result by equating real and imaginary parts in (x+iy)^2=discriminant. Then you equate real and imaginary parts, and from there it’s just algebra to find x and y, after which you can substitute in (x+iy)^2, which can be rooted to finally determine the value of z. You can also convert the discriminant to exponential form to make the root easier but that’s not always an option if it’s not a round argument obviously

yeah idk what ur ona bout

SadCeliac · Jul 31, 2023

carrotsss said:
Basically you just solve it like a normal quadratic, except what is in the root is a complex number, so you will need to find the root of the complex number to determine it. To determine this root, you can either complete the square (extremely elegant and fun, but unnecessary) or find a result by equating real and imaginary parts in (x+iy)^2=discriminant. Then you equate real and imaginary parts, and from there it’s just algebra to find x and y, after which you can substitute in (x+iy)^2, which can be rooted to finally determine the value of z. You can also convert the discriminant to exponential form to make the root easier but that’s not always an option if it’s not a round argument obviously

but why do we use the discriminant

carrotsss · Jul 31, 2023

synthesisFR said:
yeah idk what ur ona bout

I’ll send you an example tomorrow I need to sleep rn sorry

carrotsss · Jul 31, 2023

Gabriel1210 said:
but why do we use the discriminant

the discriminant is what’s being rooted, and we need to find a way to root the complex number

SadCeliac · Jul 31, 2023

*psa everyone go to sleep*

synthesisFR · Jul 31, 2023

carrotsss said:
I’ll send you an example tomorrow I need to sleep rn sorry

I SACRIFIEFD MY LIFE FOR U AND THIS IS HO W UREPAY

synthesisFR · Jul 31, 2023

Someone pls

tywebb · Aug 1, 2023

synthesisFR said:
do those 2 examples

I typed my solution to your original question in post #13 at #13 .

I did the other 2 examples but using my method instead of the method in Cambridge.

$\bold{13}\boldsymbol{.}\text{ Determine the two square roots of } -4+2i.$

$\text{Suppose the two square roots are of the form }a+bi\text{ for some real numbers }a,b$

$\text{Then }a^2-b^2+2abi=-4+2i$

$\text{Hence }a^2-b^2=-4\text{ and }2ab=2\text{ and }b=\textstyle\frac{1}{a}\text{ and so }a,b\text{ have same sign.$

$\begin{aligned}\text{Now }\textstyle a^2-\frac{1}{a^2}&=-4\\a^4-1&=-4a^2\\a^4+4a^2-1&=0\\a^2&=\textstyle\frac{-4\pm\sqrt{4^2-4(1)(-1)}}{2}\\&=-2+\sqrt5\text{ rejecting }-\text{ because for real }a, a^2\ge0\end{aligned}$

$b^2=a^2+4=2+\sqrt5$

$\text{Hence }a=\pm\sqrt{-2+\sqrt5}\text{ and }b=\pm\sqrt{2+\sqrt5}\text{ respectively and so the two square roots are}$

$\pm\left(\sqrt{-2+\sqrt5}+i\sqrt{2+\sqrt5}\right)$

--------------------------------------------------------

$\bold{14}\boldsymbol{.}\text{ Solve }z^2+(4-2i)z+1=0.$

$z=\textstyle\frac{2i-4\pm\sqrt{(4-2i)^2-4(1)(1)}}{2}=i-2\pm\sqrt{2-4i}$

$\text{If }\sqrt{2-4i}=a+ib\text{ for real }a,b\text{ then}$

$2-4i=a^2-b^2+2abi\text{ and so }a^2-b^2=2\text{ and }2ab=-4$

$\therefore b=\textstyle\frac{-2}{a}\text{ so }a,b\text{ have opposite sign$

$\begin{aligned}\text{Now }\textstyle a^2-\frac{4}{a^2}&=2\\a^4-4&=2a^2\\a^4-2a^2-4&=0\\a^2&=\textstyle\frac{2\pm\sqrt{2^2-4(1)(-4)}}{2}\\&=1+\sqrt5\text{ rejecting }-\text{ since for real }a, a^2\ge0\end{aligned}$

$b^2=a^2-2=-1+\sqrt5$

$\text{Hence }a=\pm\sqrt{1+\sqrt5},b=\mp\sqrt{-1+\sqrt5}\text{ and }$

$\begin{aligned}z&=i-2\pm\left(\sqrt{1+\sqrt5}-i\sqrt{-1+\sqrt5}\right)\\&=-2+\sqrt{1+\sqrt5}+i\left(1-\sqrt{-1+\sqrt5}\right)\text{ or }-2-\sqrt{1+\sqrt5}+i\left(1+\sqrt{-1+\sqrt5}\right)\end{aligned}$

synthesisFR · Aug 1, 2023

tywebb said:
I typed my solution to your original question in post #13 at #13 .

I did the other 2 examples but using my method instead of the method in Cambridge.

$\bold{13}\boldsymbol{.}\text{ Determine the two square roots of } -4+2i.$

$\text{Suppose the two square roots are of the form }a+bi\text{ for some real numbers }a,b$

$\text{Then }a^2-b^2+2abi=-4+2i$

$\text{Hence }a^2-b^2=-4\text{ and }2ab=2\text{ and }b=\textstyle\frac{1}{a}\text{ and so }a,b\text{ have same sign.$

$\begin{aligned}\text{Now }\textstyle a^2-\frac{1}{a^2}&=-4\\a^4-1&=-4a^2\\a^4+4a^2-1&=0\\a^2&=\textstyle\frac{-4\pm\sqrt{4^2-4(1)(-1)}}{2}\\&=-2+\sqrt5\text{ rejecting }-\text{ because for real }a, a^2\ge0\end{aligned}$

$b^2=a^2+4=2+\sqrt5$

$\text{Hence }a=\pm\sqrt{-2+\sqrt5}\text{ and }b=\pm\sqrt{2+\sqrt5}\text{ respectively and so the two square roots are}$

$\pm\left(\sqrt{-2+\sqrt5}+i\sqrt{2+\sqrt5}\right)$

--------------------------------------------------------

$\bold{14}\boldsymbol{.}\text{ Solve }z^2+(4-2i)z+1=0.$

$z=\textstyle\frac{2i-4\pm\sqrt{(4-2i)^2-4(1)(1)}}{2}=i-2\pm\sqrt{2-4i}$

$\text{If }\sqrt{2-4i}=a+ib\text{ for real }a,b\text{ then}$

$2-4i=a^2-b^2+2abi\text{ and so }a^2-b^2=2\text{ and }2ab=-4$

$\therefore b=\textstyle\frac{-2}{a}\text{ so }a,b\text{ have opposite sign$

$\begin{aligned}\text{Now }a^2-\frac{4}{a^2}&=2\\a^4-4&=2a^2\\a^4-2a^2-4&=0\\a^2&=\textstyle\frac{2\pm\sqrt{2^2-4(1)(-4)}}{2}\\&=1+\sqrt5\text{ rejecting }-\text{ since for real }a, a^2\ge0\end{aligned}$

$b^2=a^2-2=-1+\sqrt5$

$\text{Hence }a=\pm\sqrt{1+\sqrt5},b=\mp\sqrt{-1+\sqrt5}\text{ and }$

$\begin{aligned}z&=i-2\pm\left(\sqrt{1+\sqrt5}-i\sqrt{-1+\sqrt5}\right)\\&=-2+\sqrt{1+\sqrt5}+i\left(1-\sqrt{-1+\sqrt5}\right)\text{ or }-2-\sqrt{1+\sqrt5}+i\left(1+\sqrt{-1+\sqrt5}\right)\end{aligned}$

Thanks !

tywebb · Aug 1, 2023

I don't like the direction Cambridge goes with this. They seem to want to lead to a more and more formulaic approach to try to make it look more sophisticated. Such false increased sophistication is not required nor encouraged in the syllabus.

The only formula I used was the quadratic formula just to make it go along a little quicker. Even that is not necessary if you complete the squares instead.

Furthermore a more formulaic approach could lead to unrecoverable errors in exams and hence is fraught with danger.

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