If P is a complex number whose modulus is 1,t | vedclass

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(B) Given $\left(\frac{1+iz}{1-iz}\right)^4 = P$ where $|P| = 1$.Let $w = \frac{1+iz}{1-iz}$. Then $w^4 = P$.Since $|P| = 1$,we have $|w|^4 = 1$,which

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real and distinct roots

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(B) Given $\left(\frac{1+iz}{1-iz}\right)^4 = P$ where $|P| = 1$.Let $w = \frac{1+iz}{1-iz}$. Then $w^4 = P$.Since $|P| = 1$,we have $|w|^4 = 1$,which implies $|w| = 1$.So,$\left|\frac{1+iz}{1-iz}\right| = 1$.This implies $|1+iz| = |1-iz|$.Let $z = x+iy$. Then $|1+i(x+iy)| = |1-i(x+iy)|$.$|1-y+ix| = |1+y-ix|$.Squaring both sides: $(1-y)^2 + x^2 = (1+y)^2 + x^2$.$1 - 2y + y^2 + x^2 = 1 + 2y + y^2 + x^2$.$-2y = 2y$ $\Rightarrow 4y = 0$ $\Rightarrow y = 0$.Thus,$z$ must be a real number. Since $z$ is real,the roots of the equation are real.

If $P$ is a complex number whose modulus is $1$,then the equation $\left(\frac{1+iz}{1-iz}\right)^4=P$ has

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