Proof by induction of Complex Numbers $(Z^n)^*=(Z^*)^n$
Hi I have asked a few questions here before and people have been really helpful, just wondering if someone could help me out with this proof by induction question.
Simply: $(Z^n)^*=(Z^*)^n$ for all positive integers $n$.
Thank you in advance
complex-numbers induction
asked Nov 7 '16 at 5:24
user2733843user2733843
265
add a comment |
Hi I have asked a few questions here before and people have been really helpful, just wondering if someone could help me out with this proof by induction question.
Simply: $(Z^n)^*=(Z^*)^n$ for all positive integers $n$.
Thank you in advance
complex-numbers induction
asked Nov 7 '16 at 5:24
user2733843user2733843
265
add a comment |
Hi I have asked a few questions here before and people have been really helpful, just wondering if someone could help me out with this proof by induction question.
Simply: $(Z^n)^*=(Z^*)^n$ for all positive integers $n$.
Thank you in advance
complex-numbers induction
asked Nov 7 '16 at 5:24
user2733843user2733843
265
Hi I have asked a few questions here before and people have been really helpful, just wondering if someone could help me out with this proof by induction question.
Simply: $(Z^n)^*=(Z^*)^n$ for all positive integers $n$.
Thank you in advance
complex-numbers induction
complex-numbers induction
asked Nov 7 '16 at 5:24
user2733843user2733843
265
asked Nov 7 '16 at 5:24
user2733843user2733843
265
asked Nov 7 '16 at 5:24
user2733843user2733843
265
asked Nov 7 '16 at 5:24
user2733843user2733843
265
asked Nov 7 '16 at 5:24
user2733843user2733843
265
265
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Let $z=x+yi$.
For $n=2$:
$$
(z^*)^2 = (x - yi)^2=x^2-y^2-2xyi = (z^2)^*.
$$
Assume that it holds for $k=n-2$, i.e.,
$$
(z^*)^{n-2} = (z^{n-2})^*.
$$
Thus, for $(z^*)^n$
$$
(z^*)^n = (z^*)^{n-2}(z^*)^2=(z^{n-2})^*(z^2)^*=(z^{n-2}z^2)^*=(z^n)^*.
$$
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $z=x+yi$.
For $n=2$:
$$
(z^*)^2 = (x - yi)^2=x^2-y^2-2xyi = (z^2)^*.
$$
Assume that it holds for $k=n-2$, i.e.,
$$
(z^*)^{n-2} = (z^{n-2})^*.
$$
Thus, for $(z^*)^n$
$$
(z^*)^n = (z^*)^{n-2}(z^*)^2=(z^{n-2})^*(z^2)^*=(z^{n-2}z^2)^*=(z^n)^*.
$$
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
add a comment |
Let $z=x+yi$.
For $n=2$:
$$
(z^*)^2 = (x - yi)^2=x^2-y^2-2xyi = (z^2)^*.
$$
Assume that it holds for $k=n-2$, i.e.,
$$
(z^*)^{n-2} = (z^{n-2})^*.
$$
Thus, for $(z^*)^n$
$$
(z^*)^n = (z^*)^{n-2}(z^*)^2=(z^{n-2})^*(z^2)^*=(z^{n-2}z^2)^*=(z^n)^*.
$$
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
add a comment |
Let $z=x+yi$.
For $n=2$:
$$
(z^*)^2 = (x - yi)^2=x^2-y^2-2xyi = (z^2)^*.
$$
Assume that it holds for $k=n-2$, i.e.,
$$
(z^*)^{n-2} = (z^{n-2})^*.
$$
Thus, for $(z^*)^n$
$$
(z^*)^n = (z^*)^{n-2}(z^*)^2=(z^{n-2})^*(z^2)^*=(z^{n-2}z^2)^*=(z^n)^*.
$$
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
Let $z=x+yi$.
For $n=2$:
$$
(z^*)^2 = (x - yi)^2=x^2-y^2-2xyi = (z^2)^*.
$$
Assume that it holds for $k=n-2$, i.e.,
$$
(z^*)^{n-2} = (z^{n-2})^*.
$$
Thus, for $(z^*)^n$
$$
(z^*)^n = (z^*)^{n-2}(z^*)^2=(z^{n-2})^*(z^2)^*=(z^{n-2}z^2)^*=(z^n)^*.
$$
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
answered Nov 7 '16 at 9:58
V. VancakV. Vancak
10.8k2926
10.8k2926
add a comment |
add a comment |
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