InternetGangster
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Hi guys. The question states - 'Find the square roots of the complex number 21 + 20i.'
Working out so far: p2 + q2i2+2pqi = 21+20i
p2-q2+2pqi= 21+20i
2pq = 20
pq=10 and p=10/q
p2-q2=21
(10/q)2-q2=21
q4+21q2-100=0
(q2+25)(q2-4)=0
q2=25 so q=+/-5.
therefore p=+/- 2. The 2 answers give z= 2+5i and z= -2-5i. However, these answers are wrong. Can someone tell me where I am going wrong? Thanks.
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Sir Cumference
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b) is correct.
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Sir Cumference
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(Original post by InternetGangster)
Hi guys. The question states - 'Find the square roots of the complex number 21 + 20i.'
Working out so far: p2 + q2i2+2pqi = 21+20i
p2-q2+2pqi= 21+20i
2pq = 20
pq=10 and p=10/q
p2-q2=21
(10/q)2-q2=21
q4+21q2-100=0
(q2+25)(q2-4)=0
q2=25 so q=+/-5.
therefore p=+/- 2. The 2 answers give z= 2+5i and z= -2-5i. However, these answers are wrong. Can someone tell me where I am going wrong? Thanks.
The bold line is wrong. From the previous line, you generate the equations:

q^2+25=0 \ \ \text{and} \ \ q^2-4=0

The first equation implies that q^2=-25 (not 25) so this equation has no real solutions and since q is real, we can discount it. The second equation: q^2-4=0 can be used to find q.
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InternetGangster
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Report Thread starter 7 years ago
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(Original post by notnek)
The bold line is wrong. From the previous line, you generate the equations:

q^2+25=0 \ \ \text{and} \ \ q^2-4=0

The first equation implies that q^2=-25 (not 25) so this equation has no real solutions and since q is real, we can discount it. The second equation: q^2-4=0 can be used to find q.
Thank you very much. I knew it would be some ridiculously obvious mistake that only someone else would be able to point out for me.:rolleyes:
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