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Square root of 3+4i |
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Answer» Firstly squaring both side compare real and imaginary then find (a2+b2)and find the value of a and b then put in +/- (a+bi) Let x+iy=√3+4iSquaring both side(x+iy)^2=3+4ix^2-y^2+2ixy=3+4iOn comparing we getx^2-y^2=3...........(i)2xy=4Now (x^2-y^2)^2=(x^2)^2+(y^2)^2-2(x^2)+2(x^2)^2(y^2)^2-2(x^2)^2(y^2)^2. {We add and subtract 2(x^2)^2(y^2)^2}Substitute the value of x^2-y^2 and 2xy,we get(3)^2=[(x^2)+(y^2)]^2-(2xy)^2[(x^2)+(y^2)]^2=25(x^2)+(y^2)=5.......(ii)Add equation (i) and (ii)On doing so we get the value of x=2√2Sub equation (i) and (ii)On doing so we get the value of y=√2Now assign them in standard form i.e.x+iy=+-(2√2+√2)Its ur answer bdi mehntt krai yrr aapne That\'s very important question |
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