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Exercise 2.2 · Q1

Q.Evaluate the following if z=5−2iz=5-2i and w=−1+3iw=-1+3i

(i) z+wz+w
(ii) z−iwz-iw
(iii) 2z+3w2z+3w
(iv) zwzw
(v) z2+2zw+w2z^2+2zw+w^2
(vi) (z+w)2(z+w)^2
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✓ Free question

Each part is a direct substitution of z=5−2iz=5-2i and w=−1+3iw=-1+3i into the given expression, simplified using i2=−1i^2=-1; parts (v) and (vi) are the same quantity computed two ways as a cross-check.

Step 1. (i) Add z+wz+w.

z+w=(5−2i)+(−1+3i)=(5−1)+(−2+3)i=4+i.z+w=(5-2i)+(-1+3i)=(5-1)+(-2+3)i=4+i.

Step 2. (ii) Compute iwiw, then z−iwz-iw.

iw=i(−1+3i)=−i+3i2=−i−3=−3−i.iw=i(-1+3i)=-i+3i^2=-i-3=-3-i.

z−iw=(5−2i)−(−3−i)=5−2i+3+i=8−i.z-iw=(5-2i)-(-3-i)=5-2i+3+i=8-i.

Step 3. (iii) Compute 2z+3w2z+3w.

2z=10−4i,3w=−3+9i.2z=10-4i,\qquad 3w=-3+9i.

2z+3w=(10−3)+(−4+9)i=7+5i.2z+3w=(10-3)+(-4+9)i=7+5i.

Step 4. (iv) Compute zwzw.

zw=(5−2i)(−1+3i)=−5+15i+2i−6i2=−5+17i+6=1+17i.zw=(5-2i)(-1+3i)=-5+15i+2i-6i^2=-5+17i+6=1+17i.

Step 5. (v) Compute z2+2zw+w2z^2+2zw+w^2 term by term.

z2=(5−2i)2=25−20i+4i2=25−20i−4=21−20i.z^2=(5-2i)^2=25-20i+4i^2=25-20i-4=21-20i.

w2=(−1+3i)2=1−6i+9i2=1−6i−9=−8−6i.w^2=(-1+3i)^2=1-6i+9i^2=1-6i-9=-8-6i.

2zw=2(1+17i)=2+34i.2zw=2(1+17i)=2+34i.

z2+2zw+w2=(21−20i)+(2+34i)+(−8−6i)=(21+2−8)+(−20+34−6)i=15+8i.z^2+2zw+w^2=(21-20i)+(2+34i)+(-8-6i)=(21+2-8)+(-20+34-6)i=15+8i.

Step 6. (vi) Compute (z+w)2(z+w)^2 directly, as a check. Since z+w=4+iz+w=4+i from Step 1,

(z+w)2=(4+i)2=16+8i+i2=16+8i−1=15+8i,(z+w)^2=(4+i)^2=16+8i+i^2=16+8i-1=15+8i,

which matches (v) exactly — as it must, since z2+2zw+w2≡(z+w)2z^2+2zw+w^2\equiv(z+w)^2 is an algebraic identity.

✓Final answer

(i) z+w=4+iz+w=4+i (ii) z−iw=8−iz-iw=8-i (iii) 2z+3w=7+5i2z+3w=7+5i (iv) zw=1+17izw=1+17i (v) z2+2zw+w2=15+8iz^2+2zw+w^2=15+8i (vi) (z+w)2=15+8i(z+w)^2=15+8i.

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