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Raising complex number to high power - Cartesian form
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up vote
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down vote
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My question is about raising a complex number to a high power, I know how to do that with De Moivre law, but i need to get the result in cartesian form, like $z=x+iy$. and without trigonometric terms.
The problem exactly is:
Write the following complex number in the following form $z=x+iy$:
$$(3-2i)^3cdot(1-i)^9$$
I searched alot the web to find an understandible explanation to solve the problem, but still i have no idea how to do that.
Is there a special formula to raise a complex number to high power?
Thanks for help!!
complex-numbers
edited Jul 23 at 17:03
Nosrati
19.4k41544
asked Jul 23 at 12:17
D.Rotnemer
11216
add a comment |Â
up vote
2
down vote
favorite
My question is about raising a complex number to a high power, I know how to do that with De Moivre law, but i need to get the result in cartesian form, like $z=x+iy$. and without trigonometric terms.
The problem exactly is:
Write the following complex number in the following form $z=x+iy$:
$$(3-2i)^3cdot(1-i)^9$$
I searched alot the web to find an understandible explanation to solve the problem, but still i have no idea how to do that.
Is there a special formula to raise a complex number to high power?
Thanks for help!!
complex-numbers
edited Jul 23 at 17:03
Nosrati
19.4k41544
asked Jul 23 at 12:17
D.Rotnemer
11216
The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case.
– gimusi
Jul 23 at 12:19
If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate.
– Shrey Joshi
Jul 23 at 12:20
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My question is about raising a complex number to a high power, I know how to do that with De Moivre law, but i need to get the result in cartesian form, like $z=x+iy$. and without trigonometric terms.
The problem exactly is:
Write the following complex number in the following form $z=x+iy$:
$$(3-2i)^3cdot(1-i)^9$$
I searched alot the web to find an understandible explanation to solve the problem, but still i have no idea how to do that.
Is there a special formula to raise a complex number to high power?
Thanks for help!!
complex-numbers
edited Jul 23 at 17:03
Nosrati
19.4k41544
asked Jul 23 at 12:17
D.Rotnemer
11216
My question is about raising a complex number to a high power, I know how to do that with De Moivre law, but i need to get the result in cartesian form, like $z=x+iy$. and without trigonometric terms.
The problem exactly is:
Write the following complex number in the following form $z=x+iy$:
$$(3-2i)^3cdot(1-i)^9$$
I searched alot the web to find an understandible explanation to solve the problem, but still i have no idea how to do that.
Is there a special formula to raise a complex number to high power?
Thanks for help!!
complex-numbers
edited Jul 23 at 17:03
Nosrati
19.4k41544
asked Jul 23 at 12:17
D.Rotnemer
11216
edited Jul 23 at 17:03
Nosrati
19.4k41544
edited Jul 23 at 17:03
Nosrati
19.4k41544
edited Jul 23 at 17:03
Nosrati
19.4k41544
19.4k41544
asked Jul 23 at 12:17
D.Rotnemer
11216
asked Jul 23 at 12:17
D.Rotnemer
11216
asked Jul 23 at 12:17
D.Rotnemer
11216
11216
The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case.
– gimusi
Jul 23 at 12:19
If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate.
– Shrey Joshi
Jul 23 at 12:20
add a comment |Â
The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case.
– gimusi
Jul 23 at 12:19
If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate.
– Shrey Joshi
Jul 23 at 12:20
The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case.
– gimusi
Jul 23 at 12:19
The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case.
– gimusi
Jul 23 at 12:19
If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate.
– Shrey Joshi
Jul 23 at 12:20
If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate.
– Shrey Joshi
Jul 23 at 12:20
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
1
down vote
accepted
If you use DeMoivre's Formula and get $r(cos x + isin x)$ then simply distribute the $r$ to get $$rcos x + (rsin x)i$$
Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.
answered Jul 23 at 12:24
RayDansh
884214
add a comment |Â
up vote
4
down vote
For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = sqrt2 e^-ipi/4.$
Hence
$$(1 - i)^9 = 2^9/2 e^-i9pi/4 = 16sqrt2 e^-ipi/4 = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that
computing $cos(3arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.
answered Jul 23 at 12:34
David K
48.2k340107
add a comment |Â
up vote
2
down vote
If we want proceed in cartesian form we need to expand
$$(3-2i)^3cdot(1-i)^9=(3-2i)cdot (3-2i)cdot(3-2i)cdot (1-i)cdot ldots cdot (1-i)$$
The best way is by exponential form
$$z=x+iy =re^ithetaimplies z^n=r^ne^intheta$$
with
- $r=sqrtx^2+y^2$
- $theta = arctan (y/x), [+pi]$ (for $xneq 0$)
answered Jul 23 at 12:20
gimusi
65.2k73583
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
add a comment |Â
up vote
2
down vote
Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then
$$(3-2i)^3(1-i)^9=16colorblue(3-2i)^2colorred(3-2i)(1-i)$$
answered Jul 23 at 12:34
Nosrati
19.4k41544
add a comment |Â
up vote
2
down vote
You can use the binomial formula and exploit simplifications.
$$(3-2i)^3=3^3-3cdot3^2cdot2,i-3cdot3cdot2^2+2^3,i=-9-46i,$$
$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$
But in general, there is no shortcut.
Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.
I don't think that the trigonometric form is of much use because
$$(sqrt13)^3(sqrt2)^9e^-i(3arctan2/3+9pi/4)$$
doesn't tell you much more than $(3-2i)^3(1-i)^9$.
answered Jul 23 at 12:36
Yves Daoust
111k665203
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you use DeMoivre's Formula and get $r(cos x + isin x)$ then simply distribute the $r$ to get $$rcos x + (rsin x)i$$
Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.
answered Jul 23 at 12:24
RayDansh
884214
add a comment |Â
up vote
1
down vote
accepted
If you use DeMoivre's Formula and get $r(cos x + isin x)$ then simply distribute the $r$ to get $$rcos x + (rsin x)i$$
Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.
answered Jul 23 at 12:24
RayDansh
884214
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you use DeMoivre's Formula and get $r(cos x + isin x)$ then simply distribute the $r$ to get $$rcos x + (rsin x)i$$
Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.
answered Jul 23 at 12:24
RayDansh
884214
If you use DeMoivre's Formula and get $r(cos x + isin x)$ then simply distribute the $r$ to get $$rcos x + (rsin x)i$$
Now just evaluate the cosine and sine function and you're left with a complex number in $a+bi$ form.
answered Jul 23 at 12:24
RayDansh
884214
answered Jul 23 at 12:24
RayDansh
884214
answered Jul 23 at 12:24
RayDansh
884214
answered Jul 23 at 12:24
RayDansh
884214
884214
add a comment |Â
add a comment |Â
up vote
4
down vote
For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = sqrt2 e^-ipi/4.$
Hence
$$(1 - i)^9 = 2^9/2 e^-i9pi/4 = 16sqrt2 e^-ipi/4 = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that
computing $cos(3arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.
answered Jul 23 at 12:34
David K
48.2k340107
add a comment |Â
up vote
4
down vote
For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = sqrt2 e^-ipi/4.$
Hence
$$(1 - i)^9 = 2^9/2 e^-i9pi/4 = 16sqrt2 e^-ipi/4 = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that
computing $cos(3arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.
answered Jul 23 at 12:34
David K
48.2k340107
add a comment |Â
up vote
4
down vote
up vote
4
down vote
For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = sqrt2 e^-ipi/4.$
Hence
$$(1 - i)^9 = 2^9/2 e^-i9pi/4 = 16sqrt2 e^-ipi/4 = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that
computing $cos(3arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.
answered Jul 23 at 12:34
David K
48.2k340107
For the factor of $(1 - i)^9,$ I think the de Moivre form yields some insight, because $1 - i = sqrt2 e^-ipi/4.$
Hence
$$(1 - i)^9 = 2^9/2 e^-i9pi/4 = 16sqrt2 e^-ipi/4 = 16 - 16i.$$
For the factor of $(3-2i)^3$ I am not convinced by the other answers that
computing $cos(3arctan(-2/3))$ is simpler than just doing two complex multiplications by the algebraic method.
answered Jul 23 at 12:34
David K
48.2k340107
answered Jul 23 at 12:34
David K
48.2k340107
answered Jul 23 at 12:34
David K
48.2k340107
answered Jul 23 at 12:34
David K
48.2k340107
48.2k340107
add a comment |Â
add a comment |Â
up vote
2
down vote
If we want proceed in cartesian form we need to expand
$$(3-2i)^3cdot(1-i)^9=(3-2i)cdot (3-2i)cdot(3-2i)cdot (1-i)cdot ldots cdot (1-i)$$
The best way is by exponential form
$$z=x+iy =re^ithetaimplies z^n=r^ne^intheta$$
with
- $r=sqrtx^2+y^2$
- $theta = arctan (y/x), [+pi]$ (for $xneq 0$)
answered Jul 23 at 12:20
gimusi
65.2k73583
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
add a comment |Â
up vote
2
down vote
If we want proceed in cartesian form we need to expand
$$(3-2i)^3cdot(1-i)^9=(3-2i)cdot (3-2i)cdot(3-2i)cdot (1-i)cdot ldots cdot (1-i)$$
The best way is by exponential form
$$z=x+iy =re^ithetaimplies z^n=r^ne^intheta$$
with
- $r=sqrtx^2+y^2$
- $theta = arctan (y/x), [+pi]$ (for $xneq 0$)
answered Jul 23 at 12:20
gimusi
65.2k73583
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If we want proceed in cartesian form we need to expand
$$(3-2i)^3cdot(1-i)^9=(3-2i)cdot (3-2i)cdot(3-2i)cdot (1-i)cdot ldots cdot (1-i)$$
The best way is by exponential form
$$z=x+iy =re^ithetaimplies z^n=r^ne^intheta$$
with
- $r=sqrtx^2+y^2$
- $theta = arctan (y/x), [+pi]$ (for $xneq 0$)
answered Jul 23 at 12:20
gimusi
65.2k73583
If we want proceed in cartesian form we need to expand
$$(3-2i)^3cdot(1-i)^9=(3-2i)cdot (3-2i)cdot(3-2i)cdot (1-i)cdot ldots cdot (1-i)$$
The best way is by exponential form
$$z=x+iy =re^ithetaimplies z^n=r^ne^intheta$$
with
- $r=sqrtx^2+y^2$
- $theta = arctan (y/x), [+pi]$ (for $xneq 0$)
answered Jul 23 at 12:20
gimusi
65.2k73583
answered Jul 23 at 12:20
gimusi
65.2k73583
answered Jul 23 at 12:20
gimusi
65.2k73583
answered Jul 23 at 12:20
gimusi
65.2k73583
65.2k73583
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
add a comment |Â
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
The exponential form is not the best way. Because the solution is integer, whereas you will end-up with an ugly non-simplifiable trigonometric expression.
– Yves Daoust
Jul 23 at 12:28
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
@YvesDaoust Excluding the trigonometric form, which way would you suggest?
– gimusi
Jul 23 at 12:31
add a comment |Â
up vote
2
down vote
Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then
$$(3-2i)^3(1-i)^9=16colorblue(3-2i)^2colorred(3-2i)(1-i)$$
answered Jul 23 at 12:34
Nosrati
19.4k41544
add a comment |Â
up vote
2
down vote
Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then
$$(3-2i)^3(1-i)^9=16colorblue(3-2i)^2colorred(3-2i)(1-i)$$
answered Jul 23 at 12:34
Nosrati
19.4k41544
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then
$$(3-2i)^3(1-i)^9=16colorblue(3-2i)^2colorred(3-2i)(1-i)$$
answered Jul 23 at 12:34
Nosrati
19.4k41544
Hint: $(1-i)^2=-2i$ then $(1-i)^8=16$. then
$$(3-2i)^3(1-i)^9=16colorblue(3-2i)^2colorred(3-2i)(1-i)$$
answered Jul 23 at 12:34
Nosrati
19.4k41544
answered Jul 23 at 12:34
Nosrati
19.4k41544
answered Jul 23 at 12:34
Nosrati
19.4k41544
answered Jul 23 at 12:34
Nosrati
19.4k41544
19.4k41544
add a comment |Â
add a comment |Â
up vote
2
down vote
You can use the binomial formula and exploit simplifications.
$$(3-2i)^3=3^3-3cdot3^2cdot2,i-3cdot3cdot2^2+2^3,i=-9-46i,$$
$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$
But in general, there is no shortcut.
Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.
I don't think that the trigonometric form is of much use because
$$(sqrt13)^3(sqrt2)^9e^-i(3arctan2/3+9pi/4)$$
doesn't tell you much more than $(3-2i)^3(1-i)^9$.
answered Jul 23 at 12:36
Yves Daoust
111k665203
add a comment |Â
up vote
2
down vote
You can use the binomial formula and exploit simplifications.
$$(3-2i)^3=3^3-3cdot3^2cdot2,i-3cdot3cdot2^2+2^3,i=-9-46i,$$
$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$
But in general, there is no shortcut.
Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.
I don't think that the trigonometric form is of much use because
$$(sqrt13)^3(sqrt2)^9e^-i(3arctan2/3+9pi/4)$$
doesn't tell you much more than $(3-2i)^3(1-i)^9$.
answered Jul 23 at 12:36
Yves Daoust
111k665203
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You can use the binomial formula and exploit simplifications.
$$(3-2i)^3=3^3-3cdot3^2cdot2,i-3cdot3cdot2^2+2^3,i=-9-46i,$$
$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$
But in general, there is no shortcut.
Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.
I don't think that the trigonometric form is of much use because
$$(sqrt13)^3(sqrt2)^9e^-i(3arctan2/3+9pi/4)$$
doesn't tell you much more than $(3-2i)^3(1-i)^9$.
answered Jul 23 at 12:36
Yves Daoust
111k665203
You can use the binomial formula and exploit simplifications.
$$(3-2i)^3=3^3-3cdot3^2cdot2,i-3cdot3cdot2^2+2^3,i=-9-46i,$$
$$(1-i)^9=(1-i)(1-i)^8=(1-i)(-2i)^4=16(1-i).$$
But in general, there is no shortcut.
Algorithmically speaking, the most efficient way to raise to a high power is by successive squarings, which is better than repeated multiples, but for moderate exponents, you won't see a big difference.
I don't think that the trigonometric form is of much use because
$$(sqrt13)^3(sqrt2)^9e^-i(3arctan2/3+9pi/4)$$
doesn't tell you much more than $(3-2i)^3(1-i)^9$.
answered Jul 23 at 12:36
Yves Daoust
111k665203
edited Jul 23 at 13:22
answered Jul 23 at 12:36
Yves Daoust
111k665203
answered Jul 23 at 12:36
Yves Daoust
111k665203
answered Jul 23 at 12:36
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
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The best way is by trigonometric or by exponential form. The alternative is by algebraic rules but it is really a less effective way in this case.
– gimusi
Jul 23 at 12:19
If you convert the quantity to polar form into a number $r(cos x + i sin x)$ then the real part is the x coordinate and imaginary part is y coordinate.
– Shrey Joshi
Jul 23 at 12:20