If x, a, b are real numbers and 1-i x/1+i x=a-i b, then a^2+b^2 is (a) 2(1+x^2…̊

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lf `(1-ix)/(1+ix)=a-ib and a^2+b^2=1` , where `a, b in R and i = sqrt-1`, then x is equal tolf `(1-ix)/(1+ix)=a-ib and a^2+b^2=1` , where `a, b in R and i = sqrt-1`, then x is equal to
Show that a real value of `x\ ` will satisfy the equation `(1-ix)/(1+ix)=a-i b\ if\ a^2+b^2=1`Show that a real value of `x\ ` will satisfy the equation `(1-ix)/(1+ix)=a-i b\ if\ a^2+b^2=1`
"Show that a real value of `x` will satisfy the equation `(1-i x)//(1+i x)=a-i b` if `a^2+b^2=1,"Show that a real value of `x` will satisfy the equation `(1-i x)//(1+i x)=a-i b` if `a^2+b^2=1,
If a,b are real and `a^(2)+b^(2)=1,` then show that, the equation `(1-ix)/(1+ix)=a-ib` is saIf a,b are real and `a^(2)+b^(2)=1,` then show that, the equation `(1-ix)/(1+ix)=a-ib` is sa
How to solve the most hardest equation in the world || x+1=xHow to solve the most hardest equation in the world || x+1=x
If \( \frac{1-i x}{1+i x}=a+i b \), then \( a^{2}+b^{2}= \) (A) Content not found (B) \( -1 \) (...If \( \frac{1-i x}{1+i x}=a+i b \), then \( a^{2}+b^{2}= \) (A) Content not found (B) \( -1 \) (...
If a, b, x are real numbers such that ((1-ix)/(1+ix))=a-ib thenIf a, b, x are real numbers such that ((1-ix)/(1+ix))=a-ib then
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