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find the integral solutions of the equation (1-i)^x=(2)^x
Dear Student, Taking Log on both sides,u get xlog1/logi = xlog2 U have log1=0 and log2 is not equal to 0,which means x has o be zero and that is the only integral solution THANKS ALL THE BEST PLZZZ APPROVE MY ANSWER IF YOU LIKE IT
Dear Student,
Taking Log on both sides,u get xlog1/logi = xlog2
U have log1=0 and log2 is not equal to 0,which means x has o be zero
and that is the only integral solution
THANKS
ALL THE BEST
PLZZZ APPROVE MY ANSWER IF YOU LIKE IT
Dear Student, Sorry for the wrong reply,i went wrong somewhere... Anyways,taking Log on both sides,u get xLog(1-i) = x Log2 multiply and divide 1-i by root 2,u get 1-i/2^0.5 = e^-ipie/4 that is x( 0.5Log2 - i22/28 ) = xLog2 that is xLog2 = -i22/28 and x=0 are the solutions clearly,x is imaginary if the 1st eqn is true,hence,x=0 is the only integral solution THANKS ALL THE BEST PLZZZ APPROVE MY ASNWER IF YOU LIKE IT
Sorry for the wrong reply,i went wrong somewhere...
Anyways,taking Log on both sides,u get xLog(1-i) = x Log2
multiply and divide 1-i by root 2,u get 1-i/2^0.5 = e^-ipie/4
that is x( 0.5Log2 - i22/28 ) = xLog2
that is xLog2 = -i22/28 and x=0 are the solutions
clearly,x is imaginary if the 1st eqn is true,hence,x=0 is the only integral solution
PLZZZ APPROVE MY ASNWER IF YOU LIKE IT
(1-i)^x=2^x1-iComplex number is of the from x+iyWhere x=1 Y=-1I 1-i l=√x^2+y^2l 1-i l=√2 Hence l 1-i l^x=2^x Putting value (√2)^x=2^x √2^x=(√2×√2)^x √2^x=(√2)^2x Comparing powers x=2x x-2x=0 -x=0 X=0
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