Solveeit Logo
Login

Question

Mathematics Question on complex numbers

If z=x−iyz = x - iy and z13=a+ibz^{\frac{1}{3}} = a + ib, then (xa+yb)a2+b2=\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^{2}+b^{2}} =

A

-2

B

-1

C

1

D

2

Answer

-2

Explanation

Solution

Given,
z=x−iyz =x-i y and z13=a+ibz^{\frac{1}{3}}=a+i b
z1/3=a+ibz^{1 / 3} =a+i b
Take cube on both sides, we get
(z1/3)3=(a+ib)3\left(z^{1 / 3}\right)^{3}=(a+i b)^{3}
⇒z=a3+(ib)3+3a2ib+3a(ib)2\Rightarrow z=a^{3}+(i b)^{3}+3 a^{2} i b+3 a(i b)^{2}
⇒z=a3+i3b3+3a2ib+3ab2i2\Rightarrow z=a^{3}+i^{3} b^{3}+3 a^{2} i b+3 a b^{2} i^{2}
⇒z=a3−ib3+3a2bi−3ab2\Rightarrow z=a^{3}-i b^{3}+3 a^{2} b i-3 a b^{2}
[∵i2=−1]\left[\because i^{2}=-1\right]
⇒z=(a3−3ab2)−i(b3−3a2b)\Rightarrow z=\left(a^{3}-3 a b^{2}\right)-i\left(b^{3}-3 a^{2} b\right)
⇒x−iy=(a3−3ab2)−i(b3−3a2b)\Rightarrow x-i y=\left(a^{3}-3 a b^{2}\right)-i\left(b^{3}-3 a^{2} b\right)
[∵z=x−iy][\because z=x-i y]
Here, x=a3−3ab2x=a^{3}-3 a b^{2} and y=b3−3a2by=b^{3}-3 a^{2} b
Now, (xa+yb)a2+b2=(a3−3ab2a+b3−3a2bb)a2+b2\frac{\left(\frac{x}{a}+\frac{y}{b}\right)}{a^{2}+b^{2}} =\frac{\left(\frac{a^{3}-3 a b^{2}}{a}+\frac{b^{3}-3 a^{2} b}{b}\right)}{a^{2}+b^{2}}
=a2−3b2+b2−3a2a2+b2=\frac{a^{2}-3 b^{2}+b^{2}-3 a^{2}}{a^{2}+b^{2}}
=−2a2−2b2a2+b2=−2(a2+b2)a2+b2=−2=\frac{-2 a^{2}-2 b^{2}}{a^{2}+b^{2}}=\frac{-2\left(a^{2}+b^{2}\right)}{a^{2}+b^{2}}=-2