1+i=a+bi
1+i=(a+bi)2
1+i=(a2−b2)+2abi
Equating real and imaginary parts gives us;
1=a2−b2
1=2ab
This is a simultaneous equation and so we are going to solve by substituting;
a=2b1
1=(2b1)2−b2=4b21−b2
4b2=1−4b4
4b4+4b2−1=0
b2=2(4)−4±42−4(4)(−1)=2−1±2
b2=2−1−2 has no real roots so we take the positive value
b=±22(2−1)
a=±222−1+22−1
So finally the two possible roots of the complex number 1+i are ;
±(222−1+22−1+22(2−1)i)
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