complex number...
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Hi,
Please highlight your difficulties. Thanks.
Cheers,
Wen Shih -
Originally posted by Bigcable22:
Hi Bigcable22,
For the first question, (2 - i) will be the other solution. Since complex roots come in pairs, the third solution will have to be a real root. (z - (2 - i)) X (z - (2 + i)) X (z - a) = 0, where 'a' is the real root.
For the second question, substitute (1 - i) into the equation and show that it eventually becomes zero. Since (1 - i) is a root, another root will be (1 + i). Divide the entire equation using long division with [(z - (1 - i)) X (z - (1 + i))], and you will have a quadratic equation. Solve it using the quadratic formula to find the remaining two roots.
Please do note that I have little understanding of what is typed out here, as my younger sibling was the one who wrote out everything for me when I show her the questions, and I just typed out the parts which seem more important.
Cheers.
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- z = 2-i is the next solution. in doing so we assume first that a and b to be real numbers, hence conjugate root must occur in pairs.
- since z= 2-i and z=2+i are solutions, simply substitute both of them into the equation to obtain 2 new equations which u can solve simultaneously to obtain a and b.
- the last solution mus be real, because all coefficients of the polynomial are real and conjugate roots must occur in pair. since we have already obtain 1 pair of conjugate roots, the last root must be real.
- the next part verify just substitute 1-i in. show that LHS=RHS=0 to complete verification process.
- if this next part of the question is linked with the top part, you might want to do long division by dividing by 1-i. because if its linked it might give you the polynomial you found in the first part. otherwise if not linked, den since all coefficients are real, conj. roots will occur in pair and u can do what true heart mentions above.