B3 - The complex conjugate

B3 - The complex conjugate

Can a number and its partner reveal its real and imaginary parts separately? You will learn how conjugation changes a complex number, why conjugate products are real and nonnegative, and how conjugation interacts with addition and multiplication.

Conjugation as sign change

What is special about the pair 4+3i4+3\mathrm{i} and 4−3i4-3\mathrm{i}? Their real parts agree and their imaginary parts are opposites. Adding them cancels the imaginary terms; multiplying them also produces a real number. We give this pairing a name.

Complex conjugate

For z=x+iyz=x+\mathrm{i}y, where x,yx,y are real, the complex conjugate is z∗=x−iyz^*=x-\mathrm{i}y. Conjugation keeps the real part and changes the sign of the imaginary part.

Read the whole imaginary coefficient, including its sign. The conjugate of 4−3i4-3\mathrm{i} is 4+3i4+3\mathrm{i}: changing the sign of −3-3 gives +3+3. A missing real or imaginary term simply has coefficient zero.

Some sources write j\mathrm j for the same imaginary unit, with j2=−1\mathrm j^2=-1. Here we consistently use i\mathrm i; the alternate letter does not introduce a second imaginary unit.

The symbol C\mathbb{C} denotes the set of complex numbers. Thus z∈Cz\in\mathbb{C} means that zz is complex; it still permits real numbers and zero.

zzz∗z^*What changes?
4+3i4+3\mathrm{i}4−3i4-3\mathrm{i}33 becomes −3-3
−4+3i-4+3\mathrm{i}−4−3i-4-3\mathrm{i}The real part stays −4-4
5i5\mathrm{i}−5i-5\mathrm{i}The real part stays zero
−7-7−7-7The imaginary part is zero
0000Both parts are zero

Applying conjugation twice restores the number: (z∗)∗=z(z^*)^*=z. Also, z=z∗z=z^* exactly when y=−yy=-y, so y=0y=0. Thus the numbers unchanged by conjugation are precisely the real numbers, including zero.

Watch out

Wrong: if z=−4+3iz=-4+3\mathrm{i}, then z∗=4−3iz^*=4-3\mathrm{i}. That is −z-z. Correct: z∗=−4−3iz^*=-4-3\mathrm{i}; only the imaginary coefficient changes sign.

Check which operation has been performed in this new case.

Sums, differences and conjugate products

A complex number and its conjugate combine in three especially useful ways. First try z=4−3iz=4-3\mathrm{i}, so z∗=4+3iz^*=4+3\mathrm{i}: the sum is 88, the difference is −6i-6\mathrm{i}, and the product is 2525.

The same cancellations occur for any real x,yx,y:

z+z∗=(x+iy)+(x−iy)=2xz−z∗=(x+iy)−(x−iy)=2iyzz∗=(x+iy)(x−iy)=x2−(iy)2=x2+y2\begin{aligned} z+z^*&=(x+\mathrm{i}y)+(x-\mathrm{i}y)=2x\\ z-z^*&=(x+\mathrm{i}y)-(x-\mathrm{i}y)=2\mathrm{i}y\\ zz^*&=(x+\mathrm{i}y)(x-\mathrm{i}y)=x^2-(\mathrm{i}y)^2=x^2+y^2 \end{aligned}

The first two identities let you extract either part without retaining the other.

Conjugate identities

For z=x+iyz=x+\mathrm{i}y with real x,yx,y,

z+z∗=2Re⁡(z),z−z∗=2iIm⁡(z),zz∗=x2+y2z+z^*=2\operatorname{Re}(z),\qquad z-z^*=2\mathrm{i}\operatorname{Im}(z),\qquad zz^*=x^2+y^2

Not in the formula booklet: learn it.

Use the identities together, keeping the distinction between an imaginary term and its coefficient.

Worked example 1: three combinations

Question. For z=4−3iz=4-3\mathrm{i}, find z+z∗z+z^*, z−z∗z-z^* and zz∗zz^*.

Plan. Write the conjugate, then add, subtract and multiply.

z∗=4+3ichange the imaginary signz+z∗=4−3i+4+3i=8cancel imaginary termsz−z∗=4−3i−4−3i=−6isubtract both termszz∗=42−(3i)2=16+9=25difference of squares\begin{aligned} z^*&=4+3\mathrm{i} && \text{change the imaginary sign}\\ z+z^*&=4-3\mathrm{i}+4+3\mathrm{i}=8 && \text{cancel imaginary terms}\\ z-z^*&=4-3\mathrm{i}-4-3\mathrm{i}=-6\mathrm{i} && \text{subtract both terms}\\ zz^*&=4^2-(3\mathrm{i})^2=16+9=25 && \text{difference of squares} \end{aligned}

Answer. 88, −6i-6\mathrm{i} and 2525, respectively.

Check. The identities give 2(4)=82(4)=8, 2i(−3)=−6i2\mathrm{i}(-3)=-6\mathrm{i} and 42+(−3)2=254^2+(-3)^2=25.

Now change just the sign of the real part. The sum changes sign, but the difference and product do not. This separates the jobs of the two coefficients.

Worked example 2: changing the real part

Question. For w=−4−3iw=-4-3\mathrm{i}, find w+w∗w+w^*, w−w∗w-w^* and ww∗ww^*.

Plan. Use the same operations with w∗=−4+3iw^*=-4+3\mathrm{i}.

w+w∗=(−4−3i)+(−4+3i)=−8add real partsw−w∗=(−4−3i)−(−4+3i)=−6iremove the bracketww∗=(−4)2−(3i)2=25square the real coefficient\begin{aligned} w+w^*&=(-4-3\mathrm{i})+(-4+3\mathrm{i})=-8 && \text{add real parts}\\ w-w^*&=(-4-3\mathrm{i})-(-4+3\mathrm{i})=-6\mathrm{i} && \text{remove the bracket}\\ ww^*&=(-4)^2-(3\mathrm{i})^2=25 && \text{square the real coefficient} \end{aligned}

Answer. −8-8, −6i-6\mathrm{i} and 2525, respectively.

Check. Expanding the product gives 16−12i+12i−9i2=2516-12\mathrm{i}+12\mathrm{i}-9\mathrm{i}^2=25. Also, adding w∗w^* back to the difference gives −6i−4+3i=w-6\mathrm{i}-4+3\mathrm{i}=w.

There is a small but useful surprise here: the sum and difference of the pair recover the original coefficients through x=(z+z∗)/2x=(z+z^*)/2 and y=(z−z∗)/(2i)y=(z-z^*)/(2\mathrm{i}).

Watch out

Wrong: z−z∗=2yz-z^*=2y. The difference is an imaginary term: z−z∗=2iyz-z^*=2\mathrm{i}y. Its imaginary part is the real number 2y2y.

Try the same three combinations with a new number.

Nonnegative products and the zero boundary

A conjugate product can never be negative, but it can be zero. For z=3iz=3\mathrm{i}, the product is (3i)(−3i)=9(3\mathrm{i})(-3\mathrm{i})=9; for z=0z=0, the product is 00.

The identity zz∗=x2+y2zz^*=x^2+y^2 explains both facts. Each square is nonnegative because x,yx,y are real. A sum of two nonnegative numbers is zero only when both are zero:

zz∗=0  ⟺  x2+y2=0  ⟺  x=y=0  ⟺  z=0zz^*=0\iff x^2+y^2=0\iff x=y=0\iff z=0

Nonnegative

A real number is nonnegative if it is greater than or equal to zero. Zero is nonnegative, but it is not positive.

Consequently, zz∗>0zz^*>0 for every nonzero complex number zz, including nonzero real numbers and nonzero purely imaginary numbers. This is why conjugation will help us divide: it can turn a nonzero complex denominator into a positive real denominator.

The condition matters in a proof. To show that zz∗zz^* is real, the imaginary terms cancelling is enough. To show that it is positive, you also need z≠0z\ne0. If zz∗=13zz^*=13, for example, that proves z≠0z\ne0, but it does not determine zz uniquely: both 2+3i2+3\mathrm{i} and −2+3i-2+3\mathrm{i} work.

Key idea

For every complex zz, zz∗≥0zz^*\ge0. Equality occurs exactly at z=0z=0; otherwise the product is positive.

Use the boundary case to judge the following statements.

Conjugating sums and products

Conjugation can be performed before or after addition and multiplication, with the same result. For instance, (1+2i)+3=4+2i(1+2\mathrm{i})+3=4+2\mathrm{i}, whose conjugate is 4−2i4-2\mathrm{i}; adding 1−2i1-2\mathrm{i} and 33 gives that same number.

Conjugating operations

For all complex numbers u,vu,v,

(u+v)∗=u∗+v∗,(u−v)∗=u∗−v∗,(uv)∗=u∗v∗(u+v)^*=u^*+v^*,\qquad (u-v)^*=u^*-v^*,\qquad (uv)^*=u^*v^*

Not in the formula booklet: learn it.

Here is why the product rule works. Write u=a+ibu=a+\mathrm{i}b and v=c+idv=c+\mathrm{i}d, where all four coefficients are real. Then

uv=(ac−bd)+i(ad+bc)(uv)∗=(ac−bd)−i(ad+bc)u∗v∗=(a−ib)(c−id)=(ac−bd)−i(ad+bc)\begin{aligned} uv&=(ac-bd)+\mathrm{i}(ad+bc)\\ (uv)^*&=(ac-bd)-\mathrm{i}(ad+bc)\\ u^*v^*&=(a-\mathrm{i}b)(c-\mathrm{i}d)=(ac-bd)-\mathrm{i}(ad+bc) \end{aligned}

The last two expressions agree. For sums and differences, collecting the real and imaginary coefficients proves the rules in the same way. A real factor stays fixed: if kk is real, (ku)∗=ku∗(ku)^*=ku^*.

Worked example 3: conjugating a sum

Question. Let u=2+iu=2+\mathrm{i} and v=−1+3iv=-1+3\mathrm{i}. Find (u+v)∗(u+v)^*.

Plan. Add first, then change the imaginary sign.

u+v=(2−1)+(1+3)i=1+4icollect like terms(u+v)∗=1−4iconjugate the result\begin{aligned} u+v&=(2-1)+(1+3)\mathrm{i}=1+4\mathrm{i} && \text{collect like terms}\\ (u+v)^*&=1-4\mathrm{i} && \text{conjugate the result} \end{aligned}

Answer. 1−4i1-4\mathrm{i}.

Check. Conjugate separately: u∗+v∗=(2−i)+(−1−3i)=1−4iu^*+v^*=(2-\mathrm{i})+(-1-3\mathrm{i})=1-4\mathrm{i}.

Keep the same numbers and change the operation to multiplication. The rule still works, but the real part now contains the product of the imaginary terms.

Worked example 4: conjugating a product

Question. For the same u,vu,v, find (uv)∗(uv)^*.

Plan. Expand the product, simplify i2\mathrm{i}^2, then conjugate.

uv=(2+i)(−1+3i)substitute=−2+6i−i+3i2=−5+5iexpand and simplify(uv)∗=−5−5ikeep the real part\begin{aligned} uv&=(2+\mathrm{i})(-1+3\mathrm{i}) && \text{substitute}\\ &=-2+6\mathrm{i}-\mathrm{i}+3\mathrm{i}^2=-5+5\mathrm{i} && \text{expand and simplify}\\ (uv)^*&=-5-5\mathrm{i} && \text{keep the real part} \end{aligned}

Answer. −5−5i-5-5\mathrm{i}.

Check. The other route gives u∗v∗=(2−i)(−1−3i)=−2−6i+i+3i2=−5−5iu^*v^*=(2-\mathrm{i})(-1-3\mathrm{i})=-2-6\mathrm{i}+\mathrm{i}+3\mathrm{i}^2=-5-5\mathrm{i}.

These rules explain why familiar algebra survives conjugation. They will also connect a polynomial equation to a second equation when its coefficients are real.

Watch out

Wrong: (uv)∗=−uv(uv)^*=-uv. Conjugation changes the imaginary part, not the whole expression. Correct: conjugate both factors, (uv)∗=u∗v∗(uv)^*=u^*v^*.

Use either route, then check with the other.

Quick Recap

  • Conjugation changes x+iyx+\mathrm{i}y to x−iyx-\mathrm{i}y; real numbers stay fixed.
  • The sum z+z∗z+z^* isolates the real part, and z−z∗z-z^* isolates the imaginary term.
  • zz∗=x2+y2≥0zz^*=x^2+y^2\ge0, with equality exactly when z=0z=0.
  • Conjugating a sum, difference or product gives the same result as conjugating its inputs first.

A number and its conjugate reveal both coefficients and turn their product into a nonnegative real number.