Question
Solve the equation
z1=233+23iz2=−233+23iz3=−3i
Alternative Form
z1≈2.598076+1.5iz2≈−2.598076+1.5iz3=−3i
Evaluate
z3=27i
Simplify
z=327i
Rewrite the complex number in polar form
More Steps

Evaluate
27i
Determine the modulus and the argument of the complex number
r=02+272θ=arctan(027)
Calculate
More Steps

Evaluate
02+272
Calculate
0+272
Add the numbers
729
Write the number in exponential form with the base of 27
272
Reduce the index of the radical and exponent with 2
27
r=27θ=arctan(027)
Substitute the given values into the formula r(cosθ+isinθ)
27(cos(2π)+isin(2π))
z=327(cos(2π)+isin(2π))
Calculate the nth roots of a complex r(cos(θ)+i×sin(θ),using nz=nr(cosnθ+2kπ+isinnθ+2kπ)
z=327×(cos(32π+2kπ)+isin(32π+2kπ))
Simplify
z=3(cos(32π+2kπ)+isin(32π+2kπ))
Since n=3,substitute k=0,1,2 into the expression
z1=3(cos(32π+2×0×π)+isin(32π+2×0×π))z2=3(cos(32π+2×1×π)+isin(32π+2×1×π))z3=3(cos(32π+2×2π)+isin(32π+2×2π))
Calculate
More Steps

Evaluate
32π+2×0×π
Any expression multiplied by 0 equals 0
32π+0
Removing 0 doesn't change the value,so remove it from the expression
32π
Rewrite the expression
2π×31
To multiply the fractions,multiply the numerators and denominators separately
2×3π
Multiply the numbers
6π
z1=3(cos(6π)+isin(6π))z2=3(cos(32π+2×1×π)+isin(32π+2×1×π))z3=3(cos(32π+2×2π)+isin(32π+2×2π))
Calculate
More Steps

Evaluate
32π+2×1×π
Multiply the terms
32π+2π
Calculate
More Steps

Evaluate
2π+2π
Reduce fractions to a common denominator
2π+22π×2
Write all numerators above the common denominator
2π+2π×2
Multiply the terms
2π+4π
Add the numbers
25π
325π
Rewrite the expression
25π×31
To multiply the fractions,multiply the numerators and denominators separately
2×35π
Multiply the numbers
65π
z1=3(cos(6π)+isin(6π))z2=3(cos(65π)+isin(65π))z3=3(cos(32π+2×2π)+isin(32π+2×2π))
Calculate
More Steps

Evaluate
32π+2×2π
Multiply the terms
32π+4π
Calculate
More Steps

Evaluate
2π+4π
Reduce fractions to a common denominator
2π+24π×2
Write all numerators above the common denominator
2π+4π×2
Multiply the terms
2π+8π
Add the numbers
29π
329π
Rewrite the expression
29π×31
Reduce the numbers
23π×1
Multiply the numbers
23π
z1=3(cos(6π)+isin(6π))z2=3(cos(65π)+isin(65π))z3=3(cos(23π)+isin(23π))
Solution
z1=233+23iz2=−233+23iz3=−3i
Alternative Form
z1≈2.598076+1.5iz2≈−2.598076+1.5iz3=−3i
Show Solution
