Square root of a complex number

? Square root of a complex numberSample methods and examplesQuestion 1Find the square roots of -3+4iSolution 1Let z=p+qia+bi=-3+4ia+bi2=-3+4ia2+2abi-b2=-3+4ia2-b2+2abi=-3+4iEquate the coefficients:1 a2-b2=-32 2ab=4b=42a=2a and a=42b=2bSubstitute b=2a into (1)a2-2a2=-3a2-4a2=-3a4-4=-3a2a4+3a2-4=0Using the quadratic formula:a2=-3±32-4×1×-42×1a2=-3±9+162a2=-3±252a2=-3±52a=±-3±52a=±-3+52, since a is reala=±22a=±1 b=2aWhen a=1, b=2When a=-1, b=-2The square roots of -3+4i are 1+2i and -1-2iTwo methods to check a solution: Square the roots to check they equal the original complex number i.e. Show: 1+2i2=1+2i1+2i=-3+4i and -1-2i2=-1-2i-1-2i=-3+4iCheck by graphing the two simultaneous equations using graphing software.For a2-b2=-3, graph y2-x2=-3 and for 2ab=4, graph 2xy=4 and then read the points of intersection. So, x=b=2 when y=a=1 and x=b=-2 when y=a=-1Question 2Find the square root of p+qi leaving the answer in the form z=a+biSolution 2Let z=p+qia+bi=p+qia+bi2=p+qia2+2abi-b2=p+qia2-b2+2abi=p+qiEquating the coefficients:1 a2-b2=p2 2ab=qb=q2a and a=q2bSubstitute b=q2a into (1)a2-q2a2=pa2-q24a2=p4a4-q2=4pa24a4-4pa2-q2=0Using the quadratic formula:a2=--4p±-4p2-4×4×-q22×4a2=4p±16p2+16q28a2=4p±4p2+q28a2=p±p2+q22a2=Re(z)±|z|2, given the modulus of z i.e. z=p2+q2a=±Re(z)±|z|2a=±Rez+|z|2, since a is realb=q2a or b=Im(z)2a ................
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