Porisma

Worked solution

Reuse a square root before undoing a scale and translation

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

Express (1−4i)2(1-4i)^2 in the form a+bia+bi. Hence find the roots of (z2+3)2=15+8i\left(\dfrac z2+3\right)^2=15+8i. marks

[2009/MJC/I/7(a)]

CJC · Review problems · Question 7

Answer

(1−4i)2=−15−8i(1-4i)^2=-15-8i; z=2+2iz=2+2i or z=−14−2iz=-14-2i.

Square and change the sign

(1−4i)2=1−8i+16i2=−15−8i(1-4i)^2=1-8i+16i^2=-15-8i. Thus 15+8i=−(1−4i)2=[i(1−4i)]2=(4+i)215+8i=-(1-4i)^2=[i(1-4i)]^2=(4+i)^2.

How did this part go?

Choose both signs and undo the transformation

z/2+3=±(4+i)z/2+3=\pm(4+i). The positive choice gives z/2=1+iz/2=1+i, hence z=2+2iz=2+2i. The negative choice gives z/2=−7−iz/2=-7-i, hence z=−14−2iz=-14-2i. For these values the shifted quantities are 4+i4+i and −4−i-4-i, whose squares are both 15+8i15+8i.

How did this part go?

Common mistakes

The ± applies to the whole 4+i. Undo the translation and then the scale separately for each sign.

Solution notes

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