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Mathematics · Ch 10 — Complex Numbers

Subtraction of Complex Numbers

10.2.5

Subtraction of Complex Numbers

Subtraction of Complex Numbers

Let z1=a+ibz_1=a+ib and z2=c+idz_2=c+id. Subtraction is defined in terms of addition and scalar multiplication (multiplying z2z_2 by the scalar −1-1 and adding):

z1−z2=z1+(−1)z2=(a+ib)+(−c−id)=(a−c)+i(b−d)z_1-z_2=z_1+(-1)z_2=(a+ib)+(-c-id)=(a-c)+i(b-d)

So Re(z1−z2)=Re(z1)−Re(z2)\mathrm{Re}(z_1-z_2)=\mathrm{Re}(z_1)-\mathrm{Re}(z_2) and Im(z1−z2)=Im(z1)−Im(z2)\mathrm{Im}(z_1-z_2)=\mathrm{Im}(z_1)-\mathrm{Im}(z_2).

Worked examples.

  1. z1=4+3i, z2=2+iz_1=4+3i,\ z_2=2+i: z1−z2=(4+3i)−(2+i)=(4−2)+(3−1)i=2+2iz_1-z_2=(4+3i)-(2+i)=(4-2)+(3-1)i=2+2i.
  2. z1=7+i, z2=4i, z3=−3+2iz_1=7+i,\ z_2=4i,\ z_3=-3+2i: then 2z1−(5z2+2z3)=2(7+i)−[5(4i)+2(−3+2i)]=(14+2i)−[20i−6+4i]=(14+2i)−[−6+24i]=14+2i+6−24i=20−22i2z_1-(5z_2+2z_3)=2(7+i)-[5(4i)+2(-3+2i)]=(14+2i)-[20i-6+4i]=(14+2i)-[-6+24i]=14+2i+6-24i=20-22i. This example shows subtraction combined with scalar multiplication across several terms — work inside the brackets first, then subtract the whole bracket.

Properties of subtraction. …

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