COMPLEX NUMBERS · HL
Complex Equations
Square roots, quadratics & cubics.
Section 1 of 3
Square Roots
To find the square root of a complex number you cannot split the root.
$\sqrt{3+4i}$ ≠ $\sqrt{3}+\sqrt{4i}$
$\sqrt{9+16}$ ≠ $\sqrt{9}+\sqrt{16}$
Same idea — the root of a sum is not the sum of the roots.
Instead, let the answer be a complex number and square both sides.
(i) Find $\sqrt{3+4i}$
Learn
→Let $\sqrt{3+4i}=x+yi$
Square both sides: $3+4i=(x+yi)^2$
$3+4i=x^2+2xyi+y^2i^2$ ($i^2=-1$)
$3+4i=x^2+2xyi-y^2$
Now compare Real with Real, and Imaginary with Imaginary.
Real: $3=x^2-y^2$
Imaginary: $4=2xy$
From the imaginary part: $2=xy$, so $x=\dfrac{2}{y}$
Sub into the real equation:
$3=\left(\dfrac{2}{y}\right)^2-y^2=\dfrac{4}{y^2}-y^2$
Recall $\left(\dfrac{a}{b}\right)^2=\dfrac{a^2}{b^2}$.
Multiply across by $y^2$: $3y^2=4-y^4$
Let $t=y^2$ to make a quadratic:
$3t=4-t^2 \;\Rightarrow\; t^2+3t-4=0$
$(t+4)(t-1)=0$
$t=-4$ or $t=1$
$y^2=-4$ rejected — a square can't be negative.
$y^2=1$
$y=\pm 1$
If $y=1$: $x=\dfrac{2}{1}=2$. If $y=-1$: $x=\dfrac{2}{-1}=-2$.
$\sqrt{3+4i}=2+i$ or $-2-i$
The two roots are negatives of each other — opposite points through the origin.
YOU TRY · 1
Find $\sqrt{5+12i}$.
Let $\sqrt{5+12i}=x+yi$ and square both sides.
$5=x^2-y^2$, $12=2xy \Rightarrow x=\dfrac{6}{y}$. Solve: $y=\pm 2$, $x=\pm 3$.
$3+2i$ or $-3-2i$
YOU TRY · 2
Find $\sqrt{-8+6i}$.
Same method: $-8=x^2-y^2$, $6=2xy$.
$3=xy \Rightarrow x=\dfrac{3}{y}$. Solve: $y=\pm 3$, $x=\pm 1$.
$1+3i$ or $-1-3i$
Section 2 of 3
Quadratics
Three types come up. Know which method each one needs.
(i) Type 1 — a root is given, find the missing constant
$2+i$ is a root of $z^2+5z+p=0$. Find $p$.
Sub the given root in for $z$.
$z=2+i$
$(2+i)^2+5(2+i)+p=0$
$4+4i+i^2+10+5i+p=0$
$13+9i+p=0$ ($4+10-1=13$)
$p=-13-9i$
Here $p$ is complex — only one root was given, so $p$ need not be real.
(ii) Type 1 again — but with $p$ real
$z=3-5i$ is a root of $z^2-6z+p=0$, where $p\in\mathbb{R}$. Find $p$.
When $p\in\mathbb{R}$, complex roots come in conjugate pairs.
So the other root is $3+5i$.
For $z^2-(\text{sum})z+(\text{product})$, the constant $p$ is the product of the roots:
$p=(3-5i)(3+5i)=9-25i^2$
$p=9+25=34$
You could also sub $z=3-5i$ straight in — the conjugate trick is just faster.
(iii) Type 2 — form a quadratic from its roots
Learn
→$z^2-(\text{sum of roots})\,z+(\text{product of roots})=0$
Form a quadratic with roots $2\pm 3i$.
Roots: $2+3i$ and $2-3i$
Sum: $(2+3i)+(2-3i)=4$
Product: $(2+3i)(2-3i)=4-9i^2=13$
$z^2-4z+13=0$
Conjugates are mirror images in the real axis — that is why their sum and product are real.
(iv) Type 3 — solve a quadratic with a complex coefficient
Solve $2iz^2+(6+2i)z+(3-6i)=0$, where $i^2=-1$.
$a=2i$, $b=6+2i$, $c=3-6i$
$z=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(6+2i)\pm\sqrt{(6+2i)^2-4(2i)(3-6i)}}{4i}$
Work out what is under the root:
$(6+2i)^2=36+24i+4i^2=32+24i$
$-4(2i)(3-6i)=-8i(3-6i)=-24i+48i^2=-24i-48$
Add: $32+24i-24i-48=-16$
$\sqrt{-16}=\sqrt{16}\,\sqrt{-1}=4i$
$z=\dfrac{-6-2i\pm 4i}{4i}$
Split into the two roots:
$z=\dfrac{-6+2i}{4i}=\dfrac{-3+i}{2i}$ or $z=\dfrac{-6-6i}{4i}=\dfrac{-3-3i}{2i}$
Tidy each by multiplying top and bottom by $i$.
$\dfrac{-3+i}{2i}\cdot\dfrac{i}{i}=\dfrac{-3i+i^2}{2i^2}=\dfrac{-1-3i}{-2}=\dfrac{1+3i}{2}$
$\dfrac{-3-3i}{2i}\cdot\dfrac{i}{i}=\dfrac{-3i-3i^2}{2i^2}=\dfrac{3-3i}{-2}=\dfrac{-3+3i}{2}$
$z=\dfrac{1+3i}{2}$ or $z=\dfrac{-3+3i}{2}$
Quadratics — pick your method
1.Root given → sub it in.
2.Roots given → $z^2-(\text{sum})z+\text{prod}=0$.
3.Full equation → $z=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
YOU TRY · 3
$z=1-3i$ is a root of $z^2-2z+p=0$, $p\in\mathbb{R}$. Find $p$.
$p$ is real, so the other root is the conjugate.
Other root $1+3i$. $p=(1-3i)(1+3i)=1-9i^2$.
$p=10$
YOU TRY · 4
Form a real quadratic with roots $1\pm 2i$.
Find the sum and product, then $z^2-(\text{sum})z+\text{prod}=0$.
Sum $=2$. Product $=(1+2i)(1-2i)=1-4i^2=5$.
$z^2-2z+5=0$
YOU TRY · 5
Solve $iz^2+(1-i)z-1=0$.
$a=i$, $b=1-i$, $c=-1$. Use the formula and tidy with $\dfrac{i}{i}$.
$b^2-4ac=(1-i)^2+4i=2i$, $\sqrt{2i}=1+i$. $z=\dfrac{-(1-i)\pm(1+i)}{2i}$.
$z=1$ or $z=i$
Section 3 of 3
Cubics
$P(z)=z^3-kz^2+22z-20$, where $k\in\mathbb{R}$.
$3+i$ is a root of $P(z)=0$. Find $k$, then the other two roots.
(i) Find $k$
Sub $z=3+i$ in. Build up the powers first.
$(3+i)^2=9+6i+i^2=8+6i$
$(3+i)^3=(3+i)(8+6i)=24+18i+8i+6i^2=18+26i$
Sub into $P(z)=0$:
$18+26i-k(8+6i)+22(3+i)-20=0$
Group real and imaginary: $(64-8k)+(48-6k)i=0+0i$
The real part $=0$ and the imaginary part $=0$ both give the same answer.
$64-8k=0$
$k=8$
(ii) Find the other two roots
Real coefficients ⇒ complex roots come in conjugate pairs.
One root: $3+i$. Second root: $3-i$.
Sum: $(3+i)+(3-i)=6$
Product: $(3+i)(3-i)=9-i^2=10$
That conjugate pair gives the factor $z^2-6z+10$.
Divide $P(z)=z^3-8z^2+22z-20$ by $z^2-6z+10$:
$z^3-8z^2+22z-20=(z^2-6z+10)(z-2)$
Third root: $z=2$
Roots: $z=2$, $z=3+i$, $z=3-i$
The graph crosses the real axis just once (at $z=2$) — the other two roots are the complex conjugate pair, which never touch the real axis.
YOU TRY · 6
$1+2i$ is a root of $z^3-5z^2+11z-15=0$. Find the other two roots.
Real coefficients — so the conjugate is also a root. Find their factor, then divide.
Conjugate $1-2i$. Sum $2$, product $(1+2i)(1-2i)=5$ → factor $z^2-2z+5$. Divide → $z-3$.
$z=1-2i$ and $z=3$
SUM
The lot in one box
Complex equations toolkit
1.Square root: let $\sqrt{a+bi}=x+yi$, square, compare Real and Imaginary.
2.Root given, find $p$: sub the root in (or use the conjugate product when $p\in\mathbb{R}$).
3.Build a quadratic: $z^2-(\text{sum})z+\text{prod}=0$.
4.Complex coefficients: quadratic formula, then tidy with $\dfrac{i}{i}$.
5.Cubic: use the conjugate pair → factor $z^2-(\text{sum})z+\text{prod}$, then divide it out.
6.Real coefficients ⇒ complex roots always come in conjugate pairs.
End of lesson
Complex Equations — HL · Mathslive.ie