2007-01-09 01:58:24 · 8 answers · asked by pkbrauer 1 in Science & Mathematics ➔ Mathematics
5 - 6i. Always keep the real part the same (5 in this case) and negate the imaginary part (6i).
2007-01-09 02:02:21 · answer #1 · answered by Anonymous · 0⤊ 0⤋
5 -6i. The conjugate of a+bi is always a-bi, because
if a+bi is the root of a polynomial with real coefficients,
so is a -bi. Here 5 + 6i and 5-6i are the roots
of x²-10x+61=0.
2007-01-09 10:31:09 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
The conjugate of 5 + 6i =
5 - 6i
- - - - - - -s-
2007-01-09 11:51:42 · answer #3 · answered by SAMUEL D 7 · 0⤊ 0⤋
conjugate of a number of the type a+bi is a-bi
so in this case,it is 5-6i
2007-01-09 10:39:42 · answer #4 · answered by Neelu 2 · 0⤊ 0⤋
5 - 6i
2007-01-09 10:31:04 · answer #5 · answered by yodakelbell 3 · 0⤊ 0⤋
5-6i
2007-01-09 10:04:20 · answer #6 · answered by Bika_eng_egy 1 · 0⤊ 0⤋
5 - 6i
2007-01-09 10:02:22 · answer #7 · answered by Ray 5 · 0⤊ 0⤋
5-6i (conj*complex no. =real no.)
2007-01-09 10:13:57 · answer #8 · answered by Maths Rocks 4 · 0⤊ 0⤋