Complex Number flash cards

1. 1.State De Moivre's Theorem and its use for finding $n^{th}$ roots of a complex number.

   For any integer $n$, $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. For rational $n=p/q$, $\cos n\theta+i\sin n\theta$ is one of the values of $(\cos\theta+i\sin\theta)^n$. Used to derive the $n^{th}$ roots of $z=r(\cos\theta+i\sin\theta)$ as $z^{1/n}=r^{1/n}\left[\cos\frac{2k\pi+\theta}{n}+i\sin\frac{2k\pi+\theta}{n}\right]$, $k=0,1,\dots,n-1$.

   Hint: Think powers of $(\cos\theta+i\sin\theta)$; careful — theorem is exact only for integer $n$.

2. 2.What are the $n^{th}$ roots of unity and what is their sum?

   Roots of $z^n=1$ are $z=e^{i2k\pi/n}=\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}$, $k=0,1,\dots,n-1$; their sum is $0$ (and their product is $(-1)^{n-1}$).

   Hint: Sum of roots of $z^n-1=0$.

3. 3.For $z=x+iy$, define $|z|$ and $\arg(z)$, and state the key inequality relating $|z_1+z_2|$ to $|z_1|$ and $|z_2|$.

   $|z|=\sqrt{x^2+y^2}$; $\arg(z)=\tan^{-1}\left(\frac{y}{x}\right)$, adjusted for the correct quadrant of $(x,y)$. Triangle inequality: $|z_1+z_2|\leq|z_1|+|z_2|$, with equality iff $z_1,z_2$ have the same argument; also $\big||z_1|-|z_2|\big|\leq|z_1+z_2|$.

   Hint: Geometric distance from origin.

4. 4.What is the condition for four points $z_1,z_2,z_3,z_4$ (or three points) to be concyclic / collinear, and the geometric meaning of $\arg\left(\frac{z-z_1}{z-z_2}\right)$?

   Points $z_1,z_2,z_3,z_4$ are concyclic iff the cross-ratio $\dfrac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}$ is real. Geometrically, $\arg\left(\dfrac{z-z_1}{z-z_2}\right)$ is the angle subtended by segment $z_1z_2$ at the point $z$: if this angle is constant (as $z$ varies), the locus of $z$ is an arc of a circle through $z_1,z_2$; if the angle is $0$ or $\pi$, then $z_1,z_2,z$ are collinear.

   Hint: Cross-ratio real condition.

5. 5.State the relation between the cube roots of unity $1,\omega,\omega^2$ and their key algebraic properties.

   $\omega=\dfrac{-1+i\sqrt3}{2}$, $\omega^2=\dfrac{-1-i\sqrt3}{2}$, with $1+\omega+\omega^2=0$ and $\omega^3=1$; used to factor expressions like $a^3+b^3+c^3-3abc=(a+b+c)(a+\omega b+\omega^2c)(a+\omega^2b+\omega c)$.

   Hint: Non-real cube roots of $1$.

6. 6.If $z$ moves such that $|z-1|+|z+1|=4$, identify the conic traced and find its eccentricity.

   The foci are at $1$ and $-1$ so $2c=2\Rightarrow c=1$, and $2a=4\Rightarrow a=2$, giving $b^2=a^2-c^2=3$. The locus is an ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$ with eccentricity $e=\dfrac{c}{a}=\dfrac12$.

   Hint: Recognize the sum-of-distances-to-two-fixed-points definition of a conic.

7. 7.Find the locus of $z$ satisfying $|z-3|=2|z-1|$.

   Squaring, $(x-3)^2+y^2=4[(x-1)^2+y^2]$ expands to $3x^2+3y^2-2x-5=0$, i.e. $x^2+y^2-\dfrac{2}{3}x-\dfrac{5}{3}=0$. Completing the square gives the Apollonius circle centered at $\left(\dfrac13,0\right)$ with radius $\dfrac43$.

   Hint: Square both sides and complete the square — this is an Apollonius circle.

8. 8.Describe the locus of $z$ satisfying $\arg\left(\dfrac{z-1}{z+1}\right)=\dfrac{\pi}{4}$.

   This says the segment joining $1$ and $-1$ (length $2$) subtends an angle $\dfrac{\pi}{4}$ at $z$, so by the inscribed-angle relation the radius satisfies $2=2R\sin\dfrac{\pi}{4}$, giving $R=\sqrt2$. The locus is an arc of the circle of radius $\sqrt2$ through $1$ and $-1$ (only the arc on the side giving argument $+\pi/4$, not $-3\pi/4$).

   Hint: An argument condition on two fixed points always traces a circular arc.

9. 9.If $\arg(z)-\arg(z-2)=\dfrac{\pi}{2}$, what curve does $z$ trace?

   The condition says the segment from $0$ to $2$ subtends a right angle at $z$. By the converse of Thales' theorem, $z$ must lie on the circle with that segment as diameter — center $1$, radius $1$ — and specifically on the upper semicircular arc ($y>0$) that gives the stated (positive) sign of the angle, excluding the endpoints $0,2$.

   Hint: A right angle subtended by a segment is Thales' theorem territory.

10. 10.Find the locus of $z$ satisfying $|z^2-1|=|z|^2+1$.

    Write $z=x+iy$ and let $s=x^2+y^2,\,u=x^2-y^2$. Then $|z^2-1|^2=(u-1)^2+4x^2y^2=(u-1)^2+(s^2-u^2)=s^2-2u+1$, while $(|z|^2+1)^2=(s+1)^2=s^2+2s+1$. Equating gives $-2u=2s\Rightarrow u=-s\Rightarrow x^2-y^2=-(x^2+y^2)\Rightarrow x=0$. So $z$ is purely imaginary (the whole imaginary axis).

    Hint: Square both sides; use $s=x^2+y^2$, $u=x^2-y^2$ to simplify.

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