Linear Algebra Examples

Find the Square Roots of a Complex Number 4i
4i4i
Step 1
Calculate the distance from (a,b)(a,b) to the origin using the formula r=a2+b2r=a2+b2.
r=02+42r=02+42
Step 2
Simplify 02+4202+42.
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Step 2.1
Raising 00 to any positive power yields 00.
r=0+42r=0+42
Step 2.2
Raise 44 to the power of 22.
r=0+16r=0+16
Step 2.3
Add 00 and 1616.
r=16r=16
Step 2.4
Rewrite 1616 as 4242.
r=42r=42
Step 2.5
Pull terms out from under the radical, assuming positive real numbers.
r=4r=4
r=4r=4
Step 3
Calculate reference angle θ̂=arctan(|ba|)θˆ=arctan(ba).
θ̂=arctan(|40|)θˆ=arctan(40)
Step 4
The equation has an undefined fraction.
Undefined
Step 5
Since the y-coordinate is positive and the x-coordinate is 00, the point is located on y-axis between the first and fourth quadrants. The quadrants are labeled in counter-clockwise order, starting in the upper-right.
Between Quadrant 11 and 22
Step 6
Use the formula to find the roots of the complex number.
(a+bi)1n=r1ncis(θ+2πkn)(a+bi)1n=r1ncis(θ+2πkn), k=0,1,,n-1k=0,1,,n1
Step 7
Substitute rr, nn, and θθ into the formula.
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Step 7.1
Combine (4)12(4)12 and θ+2πk2θ+2πk2.
cis(4)12(θ+2πk)2cis(4)12(θ+2πk)2
Step 7.2
Combine cc and (4)12(θ+2πk)2(4)12(θ+2πk)2.
isc((4)12(θ+2πk))2isc((4)12(θ+2πk))2
Step 7.3
Combine ii and c((4)12(θ+2πk))2c((4)12(θ+2πk))2.
si(c((4)12(θ+2πk)))2si(c((4)12(θ+2πk)))2
Step 7.4
Combine ss and i(c((4)12(θ+2πk)))2i(c((4)12(θ+2πk)))2.
s(i(c((4)12(θ+2πk))))2s(i(c((4)12(θ+2πk))))2
Step 7.5
Remove parentheses.
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Step 7.5.1
Remove parentheses.
s(i(c(412(θ+2πk))))2s(i(c(412(θ+2πk))))2
Step 7.5.2
Remove parentheses.
s(i(c412(θ+2πk)))2s(i(c412(θ+2πk)))2
Step 7.5.3
Remove parentheses.
s(i(c412)(θ+2πk))2s(i(c412)(θ+2πk))2
Step 7.5.4
Remove parentheses.
s(ic412(θ+2πk))2s(ic412(θ+2πk))2
Step 7.5.5
Remove parentheses.
s(ic412)(θ+2πk)2s(ic412)(θ+2πk)2
Step 7.5.6
Remove parentheses.
s(ic)412(θ+2πk)2s(ic)412(θ+2πk)2
Step 7.5.7
Remove parentheses.
sic412(θ+2πk)2sic412(θ+2πk)2
sic412(θ+2πk)2sic412(θ+2πk)2
sic412(θ+2πk)2sic412(θ+2πk)2
Step 8
Substitute k=0k=0 into the formula and simplify.
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Step 8.1
Rewrite 44 as 2222.
k=0:(22)12cis(θ+2π(0)2)k=0:(22)12cis(θ+2π(0)2)
Step 8.2
Apply the power rule and multiply exponents, (am)n=amn(am)n=amn.
k=0:22(12)cis(θ+2π(0)2)k=0:22(12)cis(θ+2π(0)2)
Step 8.3
Cancel the common factor of 22.
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Step 8.3.1
Cancel the common factor.
k=0:22(12)cis(θ+2π(0)2)
Step 8.3.2
Rewrite the expression.
k=0:2cis(θ+2π(0)2)
k=0:2cis(θ+2π(0)2)
Step 8.4
Evaluate the exponent.
k=0:2cis(θ+2π(0)2)
Step 8.5
Multiply 2π(0).
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Step 8.5.1
Multiply 0 by 2.
k=0:2cis(θ+0π2)
Step 8.5.2
Multiply 0 by π.
k=0:2cis(θ+02)
k=0:2cis(θ+02)
k=0:2cis(θ+02)
Step 9
Substitute k=1 into the formula and simplify.
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Step 9.1
Rewrite 4 as 22.
k=1:(22)12cis(θ+2π(1)2)
Step 9.2
Apply the power rule and multiply exponents, (am)n=amn.
k=1:22(12)cis(θ+2π(1)2)
Step 9.3
Cancel the common factor of 2.
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Step 9.3.1
Cancel the common factor.
k=1:22(12)cis(θ+2π(1)2)
Step 9.3.2
Rewrite the expression.
k=1:2cis(θ+2π(1)2)
k=1:2cis(θ+2π(1)2)
Step 9.4
Evaluate the exponent.
k=1:2cis(θ+2π(1)2)
Step 9.5
Multiply 2 by 1.
k=1:2cis(θ+2π2)
k=1:2cis(θ+2π2)
Step 10
List the solutions.
k=0:2cis(θ+02)
k=1:2cis(θ+2π2)
 [x2  12  π  xdx ]