This was from tutoring but I asked both my tutor and my class teacher and they both say it's outside the scope, so I shouldn't worry about it!
I have some other questions though (from past assessments from my school).
1. Explain why multiplying a complex number z by cistheta rotates z anticlockwise about the origin, through an angle of theta.
I get the theory behind this, but I'm confused on how to set this out. Would it be best to let z= rcis(delta) and then use the multiplication of polar numbers:
so you get that: z*cistheta= rcis(delta+theta)? But I'm still a bit confused on how to prove it shifts anticlockwise
2. Assume w is a nth root of infinity. Using geometric sums, prove that w+w^2+w^3+...w^n=0
When I tried doing this, I got this: w(1-w^n)/(1-w)= w(1-w^n)/w(1/w-1)
= (1-w^n)/(1/w-1)
was I meant to use smth to do with the property 1+w+w^2=0??
3. If the mod of z=1, prove |a+bz|= |az+b|. You may assume a and b are elements of the Real numbers,.
What I did was:
square both sides: |a+bz|^2= |az+b|^2 . [[I'm just going to let the conjugates = capitalised version of the letter to make it easier to read]
(a+bz)(conjugate(a)+conjugate(bz))= (az+b)conjugate(az)+conjugate(b))
(a+bz)(A+BZ)= (az+b)(AZ+B)
aA+Abz+aBZ+BbZz=aAzZ+AZb+aBz+Bb (but Zz=1)
LHS= |a| +Abz+ aBZ+|b|
RHS= |a|+AZb+aBz +|b|
but these aren't the same? if Zz=1, isn't Z=1/z? So I'm not sure how to get the answer.
Thank you!!
If they have an answer I'm genuinely curious to see tbh, because I can't see how the coefficients work nicely when the binomial theorem is used.
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1.That is because the angle is taken with respect to the positive real axis (i.e. positive \(x\)-axis).
As the argument increases, the more we go counterclockwise, because it's always up and outwards from the positive \(x\)-axis. Or else you cannot, in some sense, make the 'angle' 'bigger'.
Clockwise shifts occur when the argument is decreased.
2. Because \(\omega\) is an nth root of
unity (watch out for the typo),
by definition it means that \( \boxed{\omega^n = 1} \). Which once you sub that in, the expression becomes 0.
3. Firstly,
never assume what you're trying to prove, or else you will receive 0 marks. You've started from the final result and worked backwards from it, but that can only be done when you
know the final result is true, not when you are trying to prove it.
To work around this issue, you should only work on one side of the equation at a time. Not both sides simultaneously.
\begin{align*} LHS&=|a+bz|^2 \\&= (a+bz)\overline{(a+bz)}\\ &= (a+bz)(a+b\overline{z}) \tag{since a and b are real}\\ &= a^2 + abz + ab\overline{z} + b^2z\overline{z}\\ &= a^2+ab(z+\overline{z})+b^2 \tag{given |z|=1} \end{align*}
\begin{align*} RHS&=|az+b|^2\\ &= (az+b)\overline{(az+b)}\\ &= (az+b)(a\overline{z}+b) \tag{again, a and b are real}\\ &= az\overline{z} + ab\overline{z}+abz+b^2\\ &= a|z|^2 +abz+ ab\overline{z} + b^2\\ &= a^2+ ab(z+\overline{z})+b^2\\ & =LHS \end{align*}
However it is worthwhile to note that yes, in general \(z^{-1} = \frac{\overline{z}}{|z|^2}\). But when \(|z|=1\), the denominator basically collapses, and we arrive at \( z^{-1}=\overline{z}\)