Z= 5 + 4i, Find square root of complex number.

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root of 41
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Easy way of dealing with the problem: 5+4i = Sq Root of (5^2+4^2) * exp (i*arctan(4/5)) Thereefore, Sq root of(5+4i) = Quadruple root of (41) * exp
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^1/2 + ^1/2]i
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well equate z=5+4i with (a+bi)^2 step1- (a+bi)^2=5+4i now expanding (a+bi)^2 we will get a^2+2abi-b^2=5+4i now in the next step--- (a^2-b^2) is a real part and 2abi is a complex part so equate it from the right side.. u will get (a^2-b^2)=5---(1) and 2abi=4i--(2) put the value of a=4/2b...
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well equate z=5+4i with (a+bi)^2 step1- (a+bi)^2=5+4i now expanding (a+bi)^2 we will get a^2+2abi-b^2=5+4i now in the next step--- (a^2-b^2) is a real part and 2abi is a complex part so equate it from the right side.. u will get (a^2-b^2)=5---(1) and 2abi=4i--(2) put the value of a=4/2b in equation ---(!) u will get (2/b)^2-b^2=5---(1) now after solving u will get value of b then put it in the equation 1 u will get value for the a then form the complex no in the form of (a+ib) that will be the answer... read less
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