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Worked solution

Real and imaginary cases with a conjugate

The complex number ww is such that kw2+kww∗+iw−iw∗−1=0kw^2+kww^*+iw-iw^*-1=0, where w∗w^* is the complex conjugate of ww and kk is a real and non-zero constant.

(i) For w=a+biw=a+bi where aa and bb are real numbers, obtain an expression for bb in terms of aa and kk. Explain why ww is either purely real or purely imaginary.

(ii) Using your result in part (i), or otherwise, find the real roots of the equation 2w2+2ww∗+iw−iw∗−1=02w^2+2ww^*+iw-iw^*-1=0.

[2016/DHS/I/2]

CJC · Discussion · Question 2

Answer

(i) b=(2ka2−1)/2b=(2ka^2-1)/2 and ab=0ab=0. (ii) w=±1/2w=\pm1/2.

(i)

Factor the source equation as kw(w+w∗)+i(w−w∗)−1=0kw(w+w^*)+i(w-w^*)-1=0. Substituting w=a+biw=a+bi, w+w∗=2aw+w^*=2a and w−w∗=2biw-w^*=2bi gives (2ka2−2b)+2abki=1(2ka^2-2b)+2abki=1. Equate real parts: 2ka2−2b=12ka^2-2b=1, so b=(2ka2−1)/2b=(2ka^2-1)/2. Equate imaginary parts: 2abk=02abk=0. Since k≠0k\ne0, ab=0ab=0, so either b=0b=0 (purely real) or a=0a=0 (purely imaginary).

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(ii) Using (i)

Real roots require b=0b=0. With k=2k=2, (4a2−1)/2=0(4a^2-1)/2=0, hence a=±1/4=±1/2a=\pm\sqrt{1/4}=\pm1/2. Thus w=−1/2w=-1/2 or w=1/2w=1/2.

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(ii) Otherwise

Directly put w=a∈Rw=a\in\mathbb R, so w∗=aw^*=a. The equation becomes 2a2+2a2+ia−ia−1=02a^2+2a^2+ia-ia-1=0, giving 4a2=14a^2=1 and the same real roots ±1/2\pm1/2.

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Common mistakes

Part (ii) asks only for real roots; do not substitute every imaginary solution into that answer.

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