Complex Numbers and Quadratic Equations — Practice

Question 1

If $|z|=1$, show that $\left|\dfrac{z-a}{1-\bar a z}ight|=1$ for any complex $a$ with $\bar a z
eq1$.

Accepted Answer

$\left|1-\bar a zight|^2=(1-\bar a z)(1-a\bar z)=1-a\bar z-\bar a z+|a|^2$.

$\left|z-aight|^2=(z-a)(\bar z-\bar a)=|z|^2-z\bar a-a\bar z+|a|^2=1-\bar a z-a\bar z+|a|^2$.

The two are identical, so the ratio of moduli is $1$.

Question 2

Find all complex numbers satisfying $z^3=\bar z$.

Accepted Answer

$|z|^3=|z|$, so $|z|=0$ or $|z|=1$.

$z=0$ works.

If $|z|=1$ then $\bar z=1/z$, so $z^3=1/z$ and $z^4=1$, giving $z=\pm1,\pm i$.

Five solutions in all, not three.

Question 3

Evaluate $\left(\dfrac{-1+i\sqrt3}{2}\right)^{2026}+\left(\dfrac{-1-i\sqrt3}{2}\right)^{2026}$.

Accepted Answer

The numbers are $\omega$ and $\omega^2$ with $\omega^3=1$.

$2026=3(675)+1$, so $\omega^{2026}=\omega$ and $\left(\omega^2ight)^{2026}=\omega^{4052}=\omega^{2}$ since $4052=3(1350)+2$.

$\omega+\omega^2=-1$.

Question 4

Find $\sqrt{-8-6i}$.

Accepted Answer

$|w|=\sqrt{64+36}=10$.

$x^2=\dfrac{10-8}{2}=1$ and $y^2=\dfrac{10+8}{2}=9$.

Since $b<0$, $x$ and $y$ have opposite signs, so the roots are $\pm(1-3i)$.

Check: $(1-3i)^2=1-6i-9=-8-6i$.

Question 5

$A(2+i)$, $B(6+4i)$ and $C$ form a triangle right-angled at $B$ with $BC=BA$ and $C$ obtained by turning $BA$ clockwise about $B$. Find $C$.

Accepted Answer

$A-B=-4-3i$.

Rotating by $-i$: $-i(-4-3i)=4i+3i^2\cdot(-1)$... carefully, $-i(-4-3i)=4i+3i^2=-3+4i$.

$C=B+(-3+4i)=3+8i$.

Check: $|C-B|=|-3+4i|=5=|A-B|$, and $(C-B)\cdot$ perpendicularity holds since $\dfrac{C-B}{A-B}=-i$ is purely imaginary.

Question 6

Find the centre and radius of the locus $2|z-i|=|z+2|$.

Accepted Answer

$4\left(x^2+(y-1)^2ight)=(x+2)^2+y^2$.

$4x^2+4y^2-8y+4=x^2+4x+4+y^2$.

$3x^2+3y^2-4x-8y=0$, that is $x^2+y^2-\dfrac43x-\dfrac83y=0$.

Centre $\left(\dfrac23,\dfrac43ight)$, radius $\sqrt{\dfrac49+\dfrac{16}{9}}=\dfrac{\sqrt{20}}{3}$.

Question 7

Describe fully the locus $\arg\dfrac{z-2}{z+2}=\dfrac{\pi}{4}$.

Accepted Answer

The segment from $-2$ to $2$ subtends $45^{\circ}$, so $z$ lies on a circle through those points with that inscribed angle. The central angle is $90^{\circ}$, so the radius is $\dfrac{4}{2\sin45^{\circ}}=2\sqrt2$ and the centre is at $(0,\pm2)$.

A positive argument places $z$ above the real axis, so the centre is $(0,-2)$ and the locus is the **major arc** above the axis, with $\pm2$ excluded.

Question 8

If $\left|z+\dfrac{1}{z}\right|=a$ with $a>0$, find the greatest value of $|z|$.

Accepted Answer

$|z|\le\left|z+\dfrac1zight|+\dfrac{1}{|z|}$, so with $r=|z|$, $r^2-ar-1\le0$ and $r\le\dfrac{a+\sqrt{a^2+4}}{2}$.

Equality needs $z$ and $1/z$ oppositely directed with $z$ real, and $r=\dfrac{a+\sqrt{a^2+4}}{2}$ satisfies $r-\dfrac1r=a$, so the bound is attained.

Question 9

For what values of $m$ does $x^2-2mx+m^2-1=0$ have both roots greater than $-2$ and less than $4$?

Accepted Answer

$(x-m)^2=1$, so the roots are $m\pm1$. $m-1>-2$ gives $m>-1$; $m+1<4$ gives $m<3$. $-10$, $f(4)=(m-3)(m-5)>0$ and $-2

Question 10

Find all $k$ for which $\left(k-1\right)x^2+2kx+k+3>0$ for every real $x$.

Accepted Answer

If $k=1$ the expression is $2x+4$, which is negative for $x<-2$, so $k=1$ fails. Otherwise we need $k-1>0$ and $D<0$: $4k^2-4(k-1)(k+3)<0$, so $4k^2-4\left(k^2+2k-3 ight)<0$, that is $-8k+12<0$ and $k>\dfrac32$. Combined with $k>1$: $k>\dfrac32$.

Question 11

If $2+i$ is a root of $x^3+ax^2+bx+10=0$ with $a$ and $b$ real, find $a$, $b$ and the third root.

Accepted Answer

$2-i$ is also a root, and their product is $5$.

The product of all three roots is $-10$, so the third root is $-2$.

Sum of roots $=(2+i)+(2-i)+(-2)=2=-a$, so $a=-2$.

Sum of pairwise products $=5+(-2)(4)=-3=b$.

Question 12

Show that $x^2+x+1=0$ and $x^2+bx+c=0$ have a common root only if $b=c=1$, given that $b$ and $c$ are real.

Accepted Answer

Subtracting gives $(1-b)x+(1-c)=0$.

If $b
eq1$ the common root is the real number $\dfrac{c-1}{1-b}$, but $x^2+x+1=0$ has no real root, so this is impossible.

Hence $b=1$, and then $1-c=0$ forces $c=1$. The two equations coincide.

Practice — Complex Numbers and Quadratic Equations

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