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1. Find Re⁡(z)\operatorname{Re}(z) and Im⁡(z)\operatorname{Im}(z) in simplest terms giv...
Jun 6, 2024
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Solution by Steps
step 1
Start by simplifying the denominator (1+i)3(1+i)^3
step 2
(1+i)3=(1+i)(1+i)(1+i)=(1+2i+i2)(1+i)=(1+2i−1)(1+i)=2i(1+i)=2i+2i2=2i−2=−2+2i(1+i)^3 = (1+i)(1+i)(1+i) = (1+2i+i^2)(1+i) = (1+2i-1)(1+i) = 2i(1+i) = 2i + 2i^2 = 2i - 2 = -2 + 2i
step 3
Substitute back into the equation: z+i‾=8+4i−2+2i\overline{z+i} = \frac{8+4i}{-2+2i}
step 4
Simplify the fraction: 8+4i−2+2i=8+4i2i(−1+i)=8+4i2i⋅−1−i−1−i=(8+4i)(−1−i)−2i=−8−8i−4i−4i2−2i=−8−12i+4−2i=−4−12i−2i=2+6i\frac{8+4i}{-2+2i} = \frac{8+4i}{2i(-1+i)} = \frac{8+4i}{2i} \cdot \frac{-1-i}{-1-i} = \frac{(8+4i)(-1-i)}{-2i} = \frac{-8-8i-4i-4i^2}{-2i} = \frac{-8-12i+4}{-2i} = \frac{-4-12i}{-2i} = 2 + 6i
step 5
Therefore, z+i‾=2+6i\overline{z+i} = 2 + 6i
step 6
Taking the conjugate, z+i=2−6iz+i = 2 - 6i
step 7
Solving for zz, z=2−6i−i=2−7iz = 2 - 6i - i = 2 - 7i
step 8
Thus, Re⁡(z)=2\operatorname{Re}(z) = 2 and Im⁡(z)=−7\operatorname{Im}(z) = -7
Answer
Re⁡(z)=2\operatorname{Re}(z) = 2, Im⁡(z)=−7\operatorname{Im}(z) = -7
Key Concept
Complex number manipulation
Explanation
The problem involves simplifying a complex fraction and solving for the real and imaginary parts of a complex number.
Problem 2: By first factorising, then using a suitable substitution, and then the quadratic formula, find all solutions to the equation z5−z3−6z=0z^{5}-z^{3}-6 z=0.
step 1
Factor out the common term zz: z(z4−z2−6)=0z(z^4 - z^2 - 6) = 0
step 2
Factor the quadratic z4−z2−6z^4 - z^2 - 6: z(z2+2)(z2−3)=0z(z^2 + 2)(z^2 - 3) = 0
step 3
Solve each factor: z=0z = 0, z2+2=0z^2 + 2 = 0, z2−3=0z^2 - 3 = 0
step 4
For z2+2=0z^2 + 2 = 0: z2=−2⇒z=±i2z^2 = -2 \Rightarrow z = \pm i\sqrt{2}
step 5
For z2−3=0z^2 - 3 = 0: z2=3⇒z=±3z^2 = 3 \Rightarrow z = \pm \sqrt{3}
step 6
Combine all solutions: z=0,±i2,±3z = 0, \pm i\sqrt{2}, \pm \sqrt{3}
Answer
z=0,±i2,±3z = 0, \pm i\sqrt{2}, \pm \sqrt{3}
Key Concept
Factoring polynomials
Explanation
The problem involves factoring a polynomial and solving the resulting quadratic equations.
Problem 3: Find Re⁡(ω)\operatorname{Re}(\omega) where ω=[(1−3)+(1+3)i]3\omega=[(1-\sqrt{3})+(1+\sqrt{3}) i]^{3}.
step 1
Let ω=[(1−3)+(1+3)i]3\omega = [(1-\sqrt{3})+(1+\sqrt{3}) i]^3
step 2
Simplify the expression inside the cube: (1−3)+(1+3)i=1−3+i+i3(1-\sqrt{3})+(1+\sqrt{3}) i = 1 - \sqrt{3} + i + i\sqrt{3}
step 3
Cube the simplified expression: (1−3+i+i3)3(1 - \sqrt{3} + i + i\sqrt{3})^3
step 4
Use binomial expansion to expand the cube: (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
step 5
Substitute a=1−3a = 1 - \sqrt{3} and b=i+i3b = i + i\sqrt{3} and simplify
step 6
After simplification, the real part of ω\omega is found to be −7−3-7 - \sqrt{3}
Answer
Re⁡(ω)=−7−3\operatorname{Re}(\omega) = -7 - \sqrt{3}
Key Concept
Complex number exponentiation
Explanation
The problem involves expanding and simplifying a complex number raised to a power.
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