4U Maths Question Thread

[RuiAce]

Thank you so much Rui

Alsoo, kinda not getting anywhere with this despite using trig formulas and whatnot:
if z is cistheta
prove that
2/(1+z) = (1+costheta+isintheta)/(1+costheta) = (2cos^2theta - 2isin(theta/2) cos(theta/2))/ 2cos^2(theta/2) = 1-itan(theta/2)

or if you can point me in the general direction? thank you!!!

\begin{align*}\frac{2}{1+z}&=\frac{2}{1+\cos\theta + i\sin \theta}\\ &= \frac{1}{\frac{1}{2}(1+\cos \theta) + \frac{1}2 i\sin \theta }\\ &= \frac{1}{\cos^2\frac\theta2 + i\cos\frac\theta2\sin\frac\theta2}\\ &= \frac{1}{\cos \frac\theta2} \times \frac{1}{\cos \frac\theta2 + i\sin \frac\theta2}\end{align*}
NOW try "realising" the denominator

[itssona]

\begin{align*}\frac{2}{1+z}&=\frac{2}{1+\cos\theta + i\sin \theta}\\ &= \frac{1}{\frac{1}{2}(1+\cos \theta) + \frac{1}2 i\sin \theta }\\ &= \frac{1}{\cos^2\frac\theta2 + i\cos\frac\theta2\sin\frac\theta2}\\ &= \frac{1}{\cos \frac\theta2} \times \frac{1}{\cos \frac\theta2 + i\sin \frac\theta2}\end{align*}
NOW try "realising" the denominator

ohh yes I got the final part!! Thank you!!
Buut wait,, how do I show that I got 1+costheta+isintheta/ 1+ costheta
?
edit: nvm, got it!!, thank you

[itssona]

hey for this, do i sub in x+iy later??

find x,y if
2z/(1+i) - 2z/i = 5/(2+i)

[itssona]

heey if my complex number has no real parts (e.g. it's in where a=0 and b=z) then how do I know my argument?

[RuiAce]

hey for this, do i sub in x+iy later??

find x,y if
2z/(1+i) - 2z/i = 5/(2+i)

\begin{align*}\frac{2z}{1+i} - \frac{2z}{i}&=\frac{5}{2+i}\\ 2z \left(\frac{1}{1+i}-\frac{1}{i}\right)&= \frac{5}{2+i}\\ 2z\left(\frac{i-(1+i)}{i(1+i)}\right)&=\frac{5}{2+i} \\ \therefore z&=\frac{5}{2(2+i)} \times \frac{i(1+i)}{-1}\end{align*}
which you should be able to finish off.

heey if my complex number has no real parts (e.g. it's in where a=0 and b=z) then how do I know my argument?

If you draw the Argand diagram, you'll see any purely imaginary number has argument either \(\frac\pi2\) or \(-\frac\pi2\)

Exercise: Work out when it's positive and when it's negative.

[itssona]

thank you so much Rui oml
also can't figure this-
E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c

so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/

[RuiAce]

thank you so much Rui oml
also can't figure this-
E+ni is a root for ax2+bx+c=0
where a and b and c are real
show that an^2 =aE^2 + bE +c

so I figured that the other root is E-ni but I've tried using product of roots, but can't seem to prove this :/

Those letters E and n look ugly so I'm gonna replace them with Greek letters........



Just use \( i^2=-1\) to finish it off from here

[itssona]

prove by Induction that z1+z2+zn = z1 +z2 +zn

where the LHS has a big conjugate line on the top and RHS has small conjugate lines (lmao)

so I proved true for n=1 and assumed for n=k but im stuck on proving n=k+1

thank you!!!
Modify message

[itssona]

z^2-2(1+i)z+8i=0
when we solve we should get (1 + root 3) +i(1-root3) and this other root, but i cant seem to get this :/

[RuiAce]

prove by Induction that z1+z2+zn = z1 +z2 +zn

where the LHS has a big conjugate line on the top and RHS has small conjugate lines (lmao)

so I proved true for n=1 and assumed for n=k but im stuck on proving n=k+1

thank you!!!
Modify message

[RuiAce]

z^2-2(1+i)z+8i=0
when we solve we should get (1 + root 3) +i(1-root3) and this other root, but i cant seem to get this :/

____________________________________




You should be able to finish it off from here.

[itssona]

thank you rui

[itssona]

i dont get this part, moduli of complex numbers removes the i?

[RuiAce]

i dont get this part, moduli of complex numbers removes the i?

\begin{align*}|(x+iy)|^2 &= |-6i|\\ |x+iy|^2 &= 6\\ x^2+y^2&=6\end{align*}

[Dragomistress]

Hello!
I would like to know if using cis is possible for complex numbers.