B2, B3 - Dividing complex numbers

B2, B3 - Dividing complex numbers

How can you divide by a number containing i\mathrm{i} and still give the answer as x+iyx+\mathrm{i}y? You will use conjugation to make denominators real, divide by any nonzero complex number, and solve linear equations by comparing real and imaginary parts.

The conjugate multiplier

What number multiplied by 2+3i2+3\mathrm{i} gives 11? Its reciprocal exists because 2+3i≠02+3\mathrm{i}\ne0. The conjugate helps us find it: (2+3i)(2−3i)=13(2+3\mathrm{i})(2-3\mathrm{i})=13, so dividing the second factor by 1313 will make the product 11.

Multiplying a fraction's numerator and denominator by the same nonzero number leaves its value unchanged. Choose the denominator's conjugate because it makes the new denominator real. This is the complex-number counterpart of rationalising a surd denominator.

Reciprocal using a conjugate

For w=a+ib≠0w=a+\mathrm{i}b\ne0, with a,ba,b real,

1w=w∗ww∗=a−iba2+b2\frac1w=\frac{w^*}{ww^*}=\frac{a-\mathrm{i}b}{a^2+b^2}

The denominator is positive because a,ba,b are not both zero.

Not in the formula booklet: learn it.

The conjugate is a convenient choice, rather than the only possible multiplier. Any nonzero real multiple of it also makes the denominator real; using the conjugate itself usually keeps the arithmetic simplest.

Worked example 1: finding a reciprocal

Question. Express 1/(2+3i)1/(2+3\mathrm{i}) in the form x+iyx+\mathrm{i}y.

Plan. Multiply the top and bottom by 2−3i2-3\mathrm{i}.

12+3i=2−3i(2+3i)(2−3i)multiply by 1=2−3i4+9use the conjugate product=213−313idivide both terms\begin{aligned} \frac1{2+3\mathrm{i}} &=\frac{2-3\mathrm{i}}{(2+3\mathrm{i})(2-3\mathrm{i})} && \text{multiply by }1\\ &=\frac{2-3\mathrm{i}}{4+9} && \text{use the conjugate product}\\ &=\frac2{13}-\frac3{13}\mathrm{i} && \text{divide both terms} \end{aligned}

Answer. 213−313i\frac2{13}-\frac3{13}\mathrm{i}.

Check. Multiplying by the original denominator gives (2+3i)(2/13−3i/13)=(4+9)/13=1(2+3\mathrm{i})(2/13-3\mathrm{i}/13)=(4+9)/13=1.

Change only the sign of the denominator's imaginary part. Its conjugate changes too, but the real denominator stays 1313.

Worked example 2: a negative imaginary coefficient

Question. Express 1/(2−3i)1/(2-3\mathrm{i}) in the form x+iyx+\mathrm{i}y.

Plan. Use 2+3i2+3\mathrm{i}, the conjugate of the whole denominator.

12−3i=2+3i(2−3i)(2+3i)multiply top and bottom=2+3i13cancel cross terms=213+313iseparate the terms\begin{aligned} \frac1{2-3\mathrm{i}} &=\frac{2+3\mathrm{i}}{(2-3\mathrm{i})(2+3\mathrm{i})} && \text{multiply top and bottom}\\ &=\frac{2+3\mathrm{i}}{13} && \text{cancel cross terms}\\ &=\frac2{13}+\frac3{13}\mathrm{i} && \text{separate the terms} \end{aligned}

Answer. 213+313i\frac2{13}+\frac3{13}\mathrm{i}.

Check. (2−3i)(2/13+3i/13)=4/13+9/13=1(2-3\mathrm{i})(2/13+3\mathrm{i}/13)=4/13+9/13=1.

The reciprocal of i\mathrm{i} is also covered by the rule: 1/i=−i1/\mathrm{i}=-\mathrm{i}, since i(−i)=1\mathrm{i}(-\mathrm{i})=1. Only division by zero is excluded.

Watch out

Wrong: 1/(2−3i)=(2+3i)/(4−9)1/(2-3\mathrm{i})=(2+3\mathrm{i})/(4-9). The product is 4−9i2=4+94-9\mathrm{i}^2=4+9, not 4−94-9.

Find a fresh reciprocal and use multiplication to check it.

Division by a nonzero complex number

A general quotient uses the same multiplier, but its numerator now needs an expansion too. For z=c+idz=c+\mathrm{i}d and w=a+ib≠0w=a+\mathrm{i}b\ne0,

zw=(c+id)(a−ib)a2+b2=ac+bda2+b2+iad−bca2+b2\begin{aligned} \frac zw &=\frac{(c+\mathrm{i}d)(a-\mathrm{i}b)}{a^2+b^2}\\ &=\frac{ac+bd}{a^2+b^2}+\mathrm{i}\frac{ad-bc}{a^2+b^2} \end{aligned}

You do not need to memorise this expanded expression. Multiplying by the conjugate and collecting terms is usually safer. Notice the subtraction in the imaginary numerator: its terms come from iad−ibc\mathrm{i}ad-\mathrm{i}bc.

Worked example 3: an exact quotient

Question. Express (4+7i)/(2+i)(4+7\mathrm{i})/(2+\mathrm{i}) in the form x+iyx+\mathrm{i}y.

Plan. Multiply top and bottom by 2−i2-\mathrm{i}, expand the top and simplify the real denominator.

4+7i2+i=(4+7i)(2−i)(2+i)(2−i)use the denominator’s conjugate=8−4i+14i−7i25expand=15+10i5=3+2idivide both terms\begin{aligned} \frac{4+7\mathrm{i}}{2+\mathrm{i}} &=\frac{(4+7\mathrm{i})(2-\mathrm{i})}{(2+\mathrm{i})(2-\mathrm{i})} && \text{use the denominator's conjugate}\\ &=\frac{8-4\mathrm{i}+14\mathrm{i}-7\mathrm{i}^2}{5} && \text{expand}\\ &=\frac{15+10\mathrm{i}}5=3+2\mathrm{i} && \text{divide both terms} \end{aligned}

Answer. 3+2i3+2\mathrm{i}.

Check. (3+2i)(2+i)=6+3i+4i+2i2=4+7i(3+2\mathrm{i})(2+\mathrm{i})=6+3\mathrm{i}+4\mathrm{i}+2\mathrm{i}^2=4+7\mathrm{i}, recovering the numerator.

Keep the numerator and change just the denominator's imaginary sign. Reusing the old multiplier would be a mistake: choose the conjugate afresh.

Worked example 4: changing the denominator

Question. Express (4+7i)/(2−i)(4+7\mathrm{i})/(2-\mathrm{i}) in the form x+iyx+\mathrm{i}y.

Plan. Use 2+i2+\mathrm{i} as the multiplier and keep exact fractions.

4+7i2−i=(4+7i)(2+i)(2−i)(2+i)conjugate the denominator=8+4i+14i+7i25expand=1+18i5=15+185iuse i2=−1\begin{aligned} \frac{4+7\mathrm{i}}{2-\mathrm{i}} &=\frac{(4+7\mathrm{i})(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})} && \text{conjugate the denominator}\\ &=\frac{8+4\mathrm{i}+14\mathrm{i}+7\mathrm{i}^2}{5} && \text{expand}\\ &=\frac{1+18\mathrm{i}}5=\frac15+\frac{18}5\mathrm{i} && \text{use }\mathrm{i}^2=-1 \end{aligned}

Answer. 15+185i\frac15+\frac{18}5\mathrm{i}.

Check. (1/5+18i/5)(2−i)=(2+18)/5+(−1+36)i/5=4+7i(1/5+18\mathrm{i}/5)(2-\mathrm{i})=(2+18)/5+(-1+36)\mathrm{i}/5=4+7\mathrm{i}.

This inverse check is stronger than checking only that the denominator is real: it checks the signs in the numerator too. It also explains why you cannot divide real and imaginary parts separately.

Watch out

Wrong: (c+id)/(a+ib)=c/a+id/b(c+\mathrm{i}d)/(a+\mathrm{i}b)=c/a+\mathrm{i}d/b. Division does not separate this way. Multiply the complete numerator and denominator by the denominator's conjugate.

Use the same method with a new numerator and denominator.

Equating real and imaginary parts

A complex equation is equivalent to two real equations: the real parts must agree and the imaginary parts must agree. Write z=x+iyz=x+\mathrm{i}y, with x,yx,y real, whenever the unknown and its conjugate occur together. Treating zz and z∗z^* as unrelated unknowns would lose their connection.

For a plain equation such as (2+i)z=4+7i(2+\mathrm{i})z=4+7\mathrm{i}, division gives z=3+2iz=3+2\mathrm{i}. Component comparison gives the same answer from 2x−y=42x-y=4 and x+2y=7x+2y=7. You can choose the shorter route.

Worked example 5: a linear equation with a conjugate

Question. Find zz if z+2z∗=6−3iz+2z^*=6-3\mathrm{i}.

Plan. Substitute z=x+iyz=x+\mathrm{i}y and z∗=x−iyz^*=x-\mathrm{i}y, then compare parts.

(x+iy)+2(x−iy)=6−3isubstitute both forms3x−iy=6−3icollect terms3x=6,−y=−3equate partsx=2,y=3solve real equations\begin{aligned} (x+\mathrm{i}y)+2(x-\mathrm{i}y)&=6-3\mathrm{i} && \text{substitute both forms}\\ 3x-\mathrm{i}y&=6-3\mathrm{i} && \text{collect terms}\\ 3x=6,\qquad -y&=-3 && \text{equate parts}\\ x=2,\qquad y&=3 && \text{solve real equations} \end{aligned}

Answer. z=2+3iz=2+3\mathrm{i}.

Check. (2+3i)+2(2−3i)=6−3i(2+3\mathrm{i})+2(2-3\mathrm{i})=6-3\mathrm{i}.

Change only the plus sign between the terms to a minus sign. Both real and imaginary coefficients are affected; expand before comparing.

Worked example 6: subtracting a conjugate term

Question. Find zz if z−2z∗=6−3iz-2z^*=6-3\mathrm{i}.

Plan. Use the same substitution, distributing the minus sign through the second bracket.

(x+iy)−2(x−iy)=6−3isubstitute−x+3iy=6−3icollect terms−x=6,3y=−3equate partsx=−6,y=−1solve\begin{aligned} (x+\mathrm{i}y)-2(x-\mathrm{i}y)&=6-3\mathrm{i} && \text{substitute}\\ -x+3\mathrm{i}y&=6-3\mathrm{i} && \text{collect terms}\\ -x=6,\qquad 3y&=-3 && \text{equate parts}\\ x=-6,\qquad y&=-1 && \text{solve} \end{aligned}

Answer. z=−6−iz=-6-\mathrm{i}.

Check. (−6−i)−2(−6+i)=6−3i(-6-\mathrm{i})-2(-6+\mathrm{i})=6-3\mathrm{i}.

The same method handles two complex unknowns. Each has two real coefficients, so two complex equations can supply four real equations.

Worked example 7: two complex unknowns

Question. The complex numbers z,wz,w satisfy z+w∗=6z+w^*=6 and 2z∗−w=3+3i2z^*-w=3+3\mathrm{i}. Find z,wz,w.

Plan. Write z=x+iyz=x+\mathrm{i}y and w=u+ivw=u+\mathrm{i}v. Separate the system into real and imaginary equations.

x+u=6,y−v=0first equation2x−u=3,−2y−v=3second equation3x=9,−3y=3add real equations; use v=yx=3,u=3,y=−1,v=−1substitute back\begin{aligned} x+u=6,\qquad y-v&=0 && \text{first equation}\\ 2x-u=3,\qquad -2y-v&=3 && \text{second equation}\\ 3x=9,\qquad -3y&=3 && \text{add real equations; use }v=y\\ x=3,\quad u=3,\quad y=-1,\quad v&=-1 && \text{substitute back} \end{aligned}

Answer. z=3−iz=3-\mathrm{i} and w=3−iw=3-\mathrm{i}.

Check. (3−i)+(3+i)=6(3-\mathrm{i})+(3+\mathrm{i})=6, and 2(3+i)−(3−i)=3+3i2(3+\mathrm{i})-(3-\mathrm{i})=3+3\mathrm{i}. Both original equations hold.

Matching both components turns complex equations into familiar real algebra. For a simultaneous system, check both original equations rather than just the final real equations.

Watch out

Wrong: 2z∗−w=2x−u+i(2y−v)2z^*-w=2x-u+\mathrm{i}(2y-v). Conjugating zz changes yy to −y-y. Correct: 2z∗−w=2x−u+i(−2y−v)2z^*-w=2x-u+\mathrm{i}(-2y-v).

Solve this new system using the same separation.

Quick Recap

  • Multiply numerator and denominator by the denominator's conjugate.
  • For a nonzero denominator a+iba+\mathrm{i}b, the new denominator a2+b2a^2+b^2 is positive.
  • Expand the whole numerator and divide both resulting terms by the real denominator.
  • Compare real and imaginary parts to solve equations, representing each conjugate consistently.

Division returns a number in the familiar form x+iyx+\mathrm{i}y, and multiplication by the original denominator checks it.