Porisma

Worked solution

Divide exactly using the conjugate of a surd denominator

A graphic calculator is not to be used in answering this question.

Two complex numbers are p=1−ip=1-i and q=−1+3 iq=-1+\sqrt3\,i, and z=p/qz=p/q. Express zz in the form x+iyx+iy, where x,yx,y are exact real values to be determined. marks

[2013/TJC/I/5(i)]

CJC · Review problems · Question 2

Answer

z=−1+34+1−34iz=-\dfrac{1+\sqrt3}{4}+\dfrac{1-\sqrt3}{4}i.

Rationalise the denominator

Multiply numerator and denominator by q∗=−1−3 iq^*=-1-\sqrt3\,i. Then (−1+3 i)(−1−3 i)=1+3=4(-1+\sqrt3\,i)(-1-\sqrt3\,i)=1+3=4.

How did this part go?

Expand and collect the two real coefficients

The numerator is (1−i)(−1−3 i)=−1−3 i+i−3=−(1+3)+(1−3)i(1-i)(-1-\sqrt3\,i)=-1-\sqrt3\,i+i-\sqrt3=-(1+\sqrt3)+(1-\sqrt3)i. Therefore z=−1+34+1−34iz=-\dfrac{1+\sqrt3}{4}+\dfrac{1-\sqrt3}{4}i. Both coefficients are exact reals; do not replace the surds by decimals.

How did this part go?

Common mistakes

The coefficient is square-root 3, confirmed on the rendered source; the text extraction loses the radical. Keep the negative contribution from i².

Solution notes

Source filename says 2026, but its printed header says 2025/JC2; both are retained without silently changing the source identity.

Identity comparison: this printed ask matches original bank 15-19 part (i), whose reviewed native record and lesson also contain the exact surd. The bank has additional parts(ii)–(iii) and a modified 2013/TJC/Prelim/I/5 attribution. This CJC occurrence retains its own 2013/TJC/I/5(i) attribution; it is not promoted or merged, and the larger bank question is unchanged.

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