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Worked solution

Reuse complex square roots in a transformed quadratic

Do not use a graphic calculator in answering this question.

(i) By letting z=a+biz=a+bi, solve z2=iz^2=i, giving your answers in exact form.

(ii) Solve w2+2w+(1−8i)=0w^2+2w+(1-8i)=0, giving the roots in Cartesian form.

(iii) Hence, solve (1−8i)v2+2iv−1=0(1-8i)v^2+2iv-1=0, giving your answers in Cartesian form.

2018/SAJC/I/5

CJC · Example 14

(i) Compare real and imaginary parts

Answer

z=12+12iz=\dfrac1{\sqrt2}+\dfrac1{\sqrt2}i or z=−12−12iz=-\dfrac1{\sqrt2}-\dfrac1{\sqrt2}i.

Let z=a+biz=a+bi with real a,ba,b. Expanding gives

(a+bi)2=a2−b2+2abi=i.\begin{aligned}&(a+bi)^2\\ & =a^2-b^2+2abi\\ & =i.\end{aligned}

Comparing components gives a2−b2=0a^2-b^2=0, so a=ba=b or a=−ba=-b, and 2ab=12ab=1, so ab=1/2ab=1/2.

For a=ba=b, a2=1/2a^2=1/2, giving a=b=1/2a=b=1/\sqrt2 or a=b=−1/2a=b=-1/\sqrt2. The signs are paired. For a=−ba=-b, a2=−1/2a^2=-1/2, impossible because aa is real. Thus the two roots are the paired values above.

How did this part go?

(ii) Reuse the square root of i

Answer

w=1+2iw=1+2i or w=−3−2iw=-3-2i.

The quadratic formula gives

\end{aligned}$$ Use the principal choice $\sqrt i=(1+i)/\sqrt2$ from (i). Its negative is already covered by the $\pm$ in the quadratic formula. Therefore $$\begin{aligned}w&=-1\pm2\sqrt2\left(\frac1{\sqrt2}+\frac1{\sqrt2}i\right)\\&=-1\pm(2+2i)\\&=1+2i\quad\text{or}\quad-3-2i.\end{aligned}$$ The selected square root is justified by (i) and the principal-argument convention; the general identity $\sqrt{u^2}=u$ is not valid for arbitrary complex $u$.
How did this part go?

(iii) Transform with w = i/v

Answer

v=25+15iv=\dfrac25+\dfrac15i or v=−213−313iv=-\dfrac2{13}-\dfrac3{13}i.

Since the constant term is −1-1, v=0v=0 cannot solve the equation. Divide by v2v^2, reorder and use i2=−1i^2=-1:

\end{aligned}$$ Compare with (ii): $w=i/v$. Hence $i/v=1+2i$ or $i/v=-3-2i$. Realise each denominator: $$\begin{aligned}v&=\frac i{1+2i}\frac{1-2i}{1-2i}=\frac{(1-2i)i}{1+2^2}=\frac{2+i}5,\\\text{or}\quad v&=\frac i{-3-2i}\frac{-3+2i}{-3+2i}=\frac{(-3+2i)i}{3^2+2^2}=\frac{-2-3i}{13}. \end{aligned}$$ Writing each fraction as separate real and imaginary terms gives the requested Cartesian form.
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