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Exercise 2.3 · Q1

Q.If z1=1−3i, z2=−4iz_1=1-3i,\ z_2=-4i, and z3=5z_3=5, show that

(i) (z1+z2)+z3=z1+(z2+z3)(z_1+z_2)+z_3=z_1+(z_2+z_3)
(ii) (z1z2)z3=z1(z2z3)(z_1z_2)z_3=z_1(z_2z_3).
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✓ Free question

We compute the left and right side of each associativity identity separately with the given numerical values and check they agree; complex numbers satisfy associativity in general because (C,+,×)(\mathbb C,+,\times) is a field, and this substitution is one concrete instance of that.

Step 1. (i) Compute the left side (z1+z2)+z3(z_1+z_2)+z_3.

z1+z2=(1−3i)+(−4i)=1−7i.z_1+z_2=(1-3i)+(-4i)=1-7i.

(z1+z2)+z3=(1−7i)+5=6−7i.(z_1+z_2)+z_3=(1-7i)+5=6-7i.

Step 2. (i) Compute the right side z1+(z2+z3)z_1+(z_2+z_3).

z2+z3=−4i+5=5−4i.z_2+z_3=-4i+5=5-4i.

z1+(z2+z3)=(1−3i)+(5−4i)=6−7i.z_1+(z_2+z_3)=(1-3i)+(5-4i)=6-7i.

Both sides equal 6−7i6-7i, so (z1+z2)+z3=z1+(z2+z3)(z_1+z_2)+z_3=z_1+(z_2+z_3) is verified.

Step 3. (ii) Compute the left side (z1z2)z3(z_1z_2)z_3.

z1z2=(1−3i)(−4i)=−4i+12i2=−4i−12=−12−4i.z_1z_2=(1-3i)(-4i)=-4i+12i^2=-4i-12=-12-4i.

(z1z2)z3=(−12−4i)(5)=−60−20i.(z_1z_2)z_3=(-12-4i)(5)=-60-20i.

Step 4. (ii) Compute the right side z1(z2z3)z_1(z_2z_3).

z2z3=(−4i)(5)=−20i.z_2z_3=(-4i)(5)=-20i.

z1(z2z3)=(1−3i)(−20i)=−20i+60i2=−20i−60=−60−20i.z_1(z_2z_3)=(1-3i)(-20i)=-20i+60i^2=-20i-60=-60-20i.

Both sides equal −60−20i-60-20i, so (z1z2)z3=z1(z2z3)(z_1z_2)z_3=z_1(z_2z_3) is verified.

Step 5. General remark. These are the associative laws for addition and multiplication in C\mathbb C; they hold for all complex numbers as a consequence of C\mathbb C being a field built from (R2,+,×)(\mathbb R^2,+,\times), not merely for this specific choice of z1,z2,z3z_1,z_2,z_3 — the substitution above just confirms the identity on a concrete case.

✓Final answer

Verified: (z1+z2)+z3=z1+(z2+z3)=6−7i(z_1+z_2)+z_3=z_1+(z_2+z_3)=6-7i, and (z1z2)z3=z1(z2z3)=−60−20i(z_1z_2)z_3=z_1(z_2z_3)=-60-20i — associativity holds for these values (and in general, since C\mathbb C is a field).

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