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Worked solution

Complex roots without a graphing calculator

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

Solve the equation z2=3−4iz^2=3-4i, giving your answers in the form a+iba+ib. Hence, by expressing 13i−14\frac13i-\frac14 in the form k(3−4i)k(3-4i), where kk is a constant, find the roots of the equation

w2=13i−14w^2=\frac13i-\frac14.

CJC · Discussion · Question 1

Answer

z=2−iz=2-i or −2+i-2+i; k=−1/12k=-1/12; w=±(1+2i)/12w=\pm(1+2i)/\sqrt{12}.

Solve for z

Let z=a+ibz=a+ib, with a,b∈Ra,b\in\mathbb R. Then (a+ib)2=(a2−b2)+2abi=3−4i(a+ib)^2=(a^2-b^2)+2abi=3-4i.

Compare real and imaginary parts: a2−b2=3a^2-b^2=3 and 2ab=−42ab=-4. Thus b=−2/ab=-2/a and

a2−4/a2=3  ⟹  (a2)2−3a2−4=0  ⟹  (a2−4)(a2+1)=0a^2-4/a^2=3\implies(a^2)^2-3a^2-4=0\implies(a^2-4)(a^2+1)=0.

Reject a2=−1a^2=-1 since a is real. The paired values are (a,b)=(2,−1)(a,b)=(2,-1) or (−2,1)(-2,1).

Therefore z=2−iz=2-i or z=−2+iz=-2+i.

How did this part go?

Hence solve for w

13i−14=−112(3−4i)\frac13i-\frac14=-\frac1{12}(3-4i), so k=−112k=-\frac1{12}.

Using (2−i)2=3−4i(2-i)^2=3-4i gives

w=±i12(2−i)=±1+2i12w=\pm\dfrac{i}{\sqrt{12}}(2-i)=\pm\dfrac{1+2i}{\sqrt{12}}.

Squaring either value returns −14+13i-\frac14+\frac13i.

How did this part go?

Common mistakes

Pair the signs of a and b using 2ab = −4; there are two roots, not four independent sign combinations.

Read the hint