Porisma

Worked solution

Practise three operations for each of three complex pairs

For each pair of complex numbers z1z_1 and z2z_2,

(a) z1=2+3iz_1=2+3i, z2=3−4iz_2=3-4i;

(b) z1=3−iz_1=3-i, z2=4−iz_2=4-i;

(c) z1=−2+5iz_1=-2+5i, z2=1−3iz_2=1-3i;

simplify (i) z1+z2z_1+z_2, (ii) z1−z2z_1-z_2 and (iii) z1z2z_1z_2.

CJC · Self-Practice 1

(a) Sum, difference and product

Answer

(i) 5−i5-i; (ii) −1+7i-1+7i; (iii) 18+i18+i.

Keep the negative imaginary part of the second number when subtracting. For the product, expand all four terms and use i2=−1i^2=-1:

z1+z2=(2+3i)+(3−4i)=5−i,z1−z2=(2+3i)−(3−4i)=−1+7i,z1z2=(2+3i)(3−4i)=6−(−12)+i(−8+(9))=18+i.\begin{aligned}z_1+z_2&=(2+3i)+(3-4i)=5-i,\\z_1-z_2&=(2+3i)-(3-4i)=-1+7i,\\z_1z_2&=(2+3i)(3-4i)\\&=6-(-12)+i(-8+(9))=18+i.\end{aligned}

How did this part go?

(b) Sum, difference and product

Answer

(i) 7−2i7-2i; (ii) −1-1; (iii) 11−7i11-7i.

The imaginary parts cancel in the difference. For the product, the two imaginary factors contribute i2=−1i^2=-1 to the real part:

z1+z2=(3−i)+(4−i)=7−2i,z1−z2=(3−i)−(4−i)=−1,z1z2=(3−i)(4−i)=12−(1)+i(−3+(−4))=11−7i.\begin{aligned}z_1+z_2&=(3-i)+(4-i)=7-2i,\\z_1-z_2&=(3-i)-(4-i)=-1,\\z_1z_2&=(3-i)(4-i)\\&=12-(1)+i(-3+(-4))=11-7i.\end{aligned}

How did this part go?

(c) Sum, difference and product

Answer

(i) −1+2i-1+2i; (ii) −3+8i-3+8i; (iii) 13+11i13+11i.

Subtract both terms of the second number, including its negative imaginary part. For the product, collect the real and imaginary contributions separately:

z1+z2=(−2+5i)+(1−3i)=−1+2i,z1−z2=(−2+5i)−(1−3i)=−3+8i,z1z2=(−2+5i)(1−3i)=−2−(−15)+i(6+(5))=13+11i.\begin{aligned}z_1+z_2&=(-2+5i)+(1-3i)=-1+2i,\\z_1-z_2&=(-2+5i)-(1-3i)=-3+8i,\\z_1z_2&=(-2+5i)(1-3i)\\&=-2-(-15)+i(6+(5))=13+11i.\end{aligned}

How did this part go?

Solution notes

All nine printed answers are retained. Three teaching parts group the three operations by the source pair; each shows all three results. Full working is authored from the printed data; the source supplies the answer table in both copies.