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Exercise 2.8 · Q2

Q.Show that (32+i2)5+(32−i2)5=−3\left(\dfrac{\sqrt3}2+\dfrac i2\right)^5+\left(\dfrac{\sqrt3}2-\dfrac i2\right)^5=-\sqrt3.

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Both bases are unit-modulus complex numbers in disguise — cis⁡(π/6)\operatorname{cis}(\pi/6) and its conjugate cis⁡(−π/6)\operatorname{cis}(-\pi/6) — so de Moivre's theorem converts the fifth powers into cis⁡(5π/6)\operatorname{cis}(5\pi/6) and cis⁡(−5π/6)\operatorname{cis}(-5\pi/6), which add to twice a cosine.

Step 1. Identify the polar form of each base. Since (32)2+(12)2=34+14=1\left(\dfrac{\sqrt3}2\right)^2+\left(\dfrac12\right)^2=\dfrac34+\dfrac14=1, both numbers have modulus 11, and

32+i2=cos⁡π6+isin⁡π6=cis⁡π6,32−i2=cos⁡π6−isin⁡π6=cis⁡(−π6).\frac{\sqrt3}2+\frac i2=\cos\frac\pi6+i\sin\frac\pi6=\operatorname{cis}\frac\pi6,\qquad \frac{\sqrt3}2-\frac i2=\cos\frac\pi6-i\sin\frac\pi6=\operatorname{cis}\left(-\frac\pi6\right).

Step 2. Raise each to the 5th power using de Moivre's theorem.

(cis⁡π6)5=cis⁡5π6,(cis⁡(−π6))5=cis⁡(−5π6).\left(\operatorname{cis}\frac\pi6\right)^5=\operatorname{cis}\frac{5\pi}6,\qquad \left(\operatorname{cis}\left(-\frac\pi6\right)\right)^5=\operatorname{cis}\left(-\frac{5\pi}6\right).

Step 3. Add the two results. Using cis⁡ϕ+cis⁡(−ϕ)=2cos⁡ϕ\operatorname{cis}\phi+\operatorname{cis}(-\phi)=2\cos\phi (Corollary, §2.8.1):

cis⁡5π6+cis⁡(−5π6)=2cos⁡5π6.\operatorname{cis}\frac{5\pi}6+\operatorname{cis}\left(-\frac{5\pi}6\right)=2\cos\frac{5\pi}6.

Step 4. Evaluate cos⁡5π6\cos\dfrac{5\pi}6. cos⁡5π6=−cos⁡π6=−32\cos\dfrac{5\pi}6=-\cos\dfrac\pi6=-\dfrac{\sqrt3}2, so

2cos⁡5π6=2(−32)=−3.2\cos\frac{5\pi}6=2\left(-\frac{\sqrt3}2\right)=-\sqrt3.

Therefore

(32+i2)5+(32−i2)5=−3.\left(\frac{\sqrt3}2+\frac i2\right)^5+\left(\frac{\sqrt3}2-\frac i2\right)^5=-\sqrt3.

✓Final answer

−3-\sqrt3, as required.

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