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Worked solution

Simultaneous equations over complex numbers

Solve the following simultaneous equations:

(a) 4iz−3w=1+5i4iz-3w=1+5i and 2z+(1+i)w=2+6i2z+(1+i)w=2+6i,

(b) iz+2w=1iz+2w=1 and 4z+(3−i)w∗=−64z+(3-i)w^*=-6,

giving zz and ww in the form a+bia+bi where aa and bb are real.

CJC · Discussion · Question 4

Answer

(a) z=5+2iz=5+2i, w=−3+5iw=-3+5i. (b) z=−2+iz=-2+i, w=1+iw=1+i.

(a)

Multiply the second equation by 2i2i: 4iz+2iw−2w=4i−124iz+2iw-2w=4i-12. Subtract the first equation: w+2iw=−13−iw+2iw=-13-i, hence (1+2i)w=−13−i(1+2i)w=-13-i. Rationalise: w=(−13−i)(1−2i)(1+2i)(1−2i)=−15+25i5=−3+5iw=\dfrac{(-13-i)(1-2i)}{(1+2i)(1-2i)}=\dfrac{-15+25i}{5}=-3+5i. Substitute into the second equation: 2z=2+6i−(1+i)(−3+5i)=2+6i−(−8+2i)=10+4i2z=2+6i-(1+i)(-3+5i)=2+6i-(-8+2i)=10+4i. Thus z=5+2iz=5+2i.

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(b)

From the first equation z=(1−2w)/i=−i+2iwz=(1-2w)/i=-i+2iw. Substitute into the second: −4i+8iw+3w∗−iw∗=−6-4i+8iw+3w^*-iw^*=-6, so 8iw+3w∗−iw∗=−6+4i8iw+3w^*-iw^*=-6+4i. Put w=a+biw=a+bi, w∗=a−biw^*=a-bi. Expansion gives (3a−9b)+(7a−3b)i=−6+4i(3a-9b)+(7a-3b)i=-6+4i. Thus 3a−9b=−63a-9b=-6 and 7a−3b=47a-3b=4. Multiply the latter by 3 and subtract the former: 18a=1818a=18. Hence a=1a=1, b=1b=1 and w=1+iw=1+i. Finally z=−i+2i(1+i)=−2+iz=-i+2i(1+i)=-2+i.

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Common mistakes

Remember 1/i=−i and keep w distinct from its conjugate.

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