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Question: If \(b + ia\), then \(a + ib\)is equal to....

If b+iab + ia, then a+iba + ibis equal to.

A

5

B

6

C

– 5

D

– 4

Answer

5

Explanation

Solution

Given that b=20b = 20⇒ 1−2i2+i+4−i3+2i=(1−2i)(3+2i)+(4−i)(2+i)(2+i)(3+2i)\frac{1 - 2i}{2 + i} + \frac{4 - i}{3 + 2i} = \frac{(1 - 2i)(3 + 2i) + (4 - i)(2 + i)}{(2 + i)(3 + 2i)},

=50−120i65=1013−2413i= \frac{50 - 120i}{65} = \frac{10}{13} - \frac{24}{13}iand a+ib>c+ida + ib > c + id

∴ib=id=0ib = id = 0

Aliter : b=d=0b = d = 0⇒ (∵i≠0)(\because i \neq 0)

⇒x+iy=32+cos⁡θ+isin⁡θx + iy = \frac{3}{2 + \cos\theta + i\sin\theta}

=3(2+cos⁡θ−isin⁡θ)(2+cos⁡θ)2+sin⁡2θ=6+3cos⁡θ−3isin⁡θ4+cos⁡2θ+4cos⁡θ+sin⁡2θ= \frac{3(2 + \cos\theta - i\sin\theta)}{(2 + \cos\theta)^{2} + \sin^{2}\theta} = \frac{6 + 3\cos\theta - 3i\sin\theta}{4 + \cos^{2}\theta + 4\cos\theta + \sin^{2}\theta}

=[6+3cos⁡θ5+4cos⁡θ]+i[−3sin⁡θ5+4cos⁡θ]= \left\lbrack \frac{6 + 3\cos\theta}{5 + 4\cos\theta} \right\rbrack + i\left\lbrack \frac{- 3\sin\theta}{5 + 4\cos\theta} \right\rbrack

a2+b2+c2+2a+a2−b2−c2+2ib(1+a)1+1+2ac+2(a+c)\frac{a^{2} + b^{2} + c^{2} + 2a + a^{2} - b^{2} - c^{2} + 2ib(1 + a)}{1 + 1 + 2ac + 2(a + c)}