√a+ib = x+iy then proof that √a-ib = x-iy

Question

It's question from complex no

 Sir please proof it step by step

Answer 1

Starting with the original equation

 

√a+ib=x+iy

 

equating the real terms of this equation note that 

 

√a=x, we'll use this relationship near the end of the proof

 

Now we'll go back to the original equation and square both sides

 

(√a+ib)²=(x+iy)²

(√a+ib)(√a+ib)=(x+iy)(x+iy)

a+2√abi+i²b²=x²+2xyi+i²y²

noting i²=-1 implies

a+2√abi-b²=x²+2xyi-y²  

 

equating the imaginary term on either side of the equation we have

 

2√ab=2xy

b=xy/√a, we'll use this relationship in just a second

 

Back to the original equation, again

 

√a+ib=x+iy

 

subtracting -2ib from either side

 

√a-ib=x+iy-2ib=x+i(y-2b)

 

Now lets substitute in the result we obtained by squaring the original equation, namely

 

b=xy/√a

 

yields

 

√a-ib=x+i(y-2xy/√a)=x+iy(1-2x/√a)

 

finally, lets substitute the intial result we got at the beginning of this proof ie √a=x

 

√a-ib=x+iy(1-2x/x)=x+iy(1-2)

 

Therefor

 

√a-ib=x-iy

 

Good Luck.

 

Cheers

 

-Chris