_Show that, if f is holomorphic and nonzero in G, then ln|f(x, y)| is harmonic in G._ Assume f = u + iv is holomorphic and nonzero in G. Since f is holomorphic, its real and imaginary parts are harmonic and therefore infinitely differentiable by Proposition 6.4 and Corollary 6.9. The first derivative of ln|f(x, y)| with respect to x is \\ (ln|f(x,y)|)_x &= [ln(+v^{2}})]_x\\ &= {2}[ln(u^{2}+v^{2})]_x\\ &={2(u^{2}+v^{2})}\\ &={u^{2}+v^{2}} And the second derivative with respect to x is (ln|f(x,y)|)_{xx}&= ({u^{2}+v^{2}})_x\\ &= +v_xv_x + vv_{xx})(u^{2}+v^{2})-(uu_x + vv_x)(2uu_x+2vv_x)}{(u^{2}+v^{2})^{2}} Similarly the first and second derivatives of ln|f(x, y)| with respect to y are (ln|f(x,y)|)_y &= [ln(+v^{2}})]_y\\ &= {2}[ln(u^{2}+v^{2})]_y\\ &={2(u^{2}+v^{2})}\\ &={u^{2}+v^{2}} and (ln|f(x,y)|)_{yy}&= ({u^{2}+v^{2}})_y\\ &= +v_yv_y + vv_{yy})(u^{2}+v^{2})-(uu_y + vv_y)(2uu_y+2vv_y)}{(u^{2}+v^{2})^{2}} Since f is nonzero and u and v are harmonic in G, the second partials of ln|f(x, y)| are defined on G. Now consider the Laplace equation, (ln|f(x,y)|)_{xx}+(ln|f(x,y)|)_{yy} &={(u^{2}+v^{2})^{2}}[(uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2})-(uu_x + vv_x)(2uu_x+2vv_x)\\ &+ (uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})-(uu_y + vv_y)(2uu_y+2vv_y)] We will show that the numerator expression will be equal to zero and thus the Laplace equation will equal zero. (uu_x + uu_{xx} +v_xv_x + vv_{xx})(u^{2}+v^{2}) &-(uu_x + vv_x)(2uu_x+2vv_x)+(uu_y + uu_{yy} +v_yv_y + vv_{yy})(u^{2}+v^{2})\\ &-(uu_y + vv_y)(2uu_y+2vv_y)\\ &=u^{2}u^{2}_x+u^{3}u_{xx}+u^{2}v^{2}_x+u^{2}vv_{xx}+v^{2}u^{2}_x+v^{2}uu_{xx}+v^{2}v^{2}_x+v^{3}v_{xx}\\ &-2u^{2}u^{2}_x-2uvv_xu_x-2uvu_xv_x-2v^{2}v^{2}_x+u^{2}u^{2}_y+u^{3}u_{yy}+u^{2}v^{2}_y+u^{2}vv_{yy}+\\ &v^{2}u^{2}_y+v^{2}uu_{yy}+v^{2}v^{2}_y+v^{3}v_{yy}-2u^{2}u^{2}_y-2uvv_yu_y-2uvu_yv_y-2v^{2}v^{2}_y\\ &-2u^{2}u^{2}_y-2uvv_yu_y-2uvu_yv_y-2v^{2}v^{2}_y But since u and v are harmonic, uxx + uyy = 0 and vxx + vyy = 0. Using this fact and combining like terms gives us, -u^{2}u^{2}_x+u^{2}v^{2}_x+v^{2}u^{2}_x-v^{2}v^{2}_x-4uvu_xv_x-u^{2}u^{2}_y+u^{2}v^{2}_y+v^{2}u^{2}_y-v^{2}v^{2}_y-4uvu_yv_y And now we can use the Cauchy-Riemann equations to get -u^{2}u^{2}_x+u^{2}v^{2}_x+v^{2}u^{2}_x-v^{2}v^{2}_x-4uvv_yv_x-u^{2}v^{2}_x+u^{2}u^{2}_x+v^{2}v^{2}_x-v^{2}u^{2}_x+4uv_xv_y Which simplifies to 0 and thus (ln|f(x, y)|)xx + (ln|f(x, y)|)yy = 0. Since ln|f(x, y)| has continuous second partials on G and the Laplace equation is satisfied, ln|f(x, y)| is harmonic on G.

_Compute_$^{\infty}{x^{4}+1}$ Let σR be the semicircle, oriented counter-clockwise, formed by the segment [ − R, R] on the real axis and the circular arc, γR, of radius R in the upper half of the complex plane. Let γ₁ be the quarter circle, oriented counter-clockwise, whose path is given by the real and imaginary axes and the portion of γR in the first quadrant. Let γ₂ be the quarter circle, oriented counter-clockwise, whose boundary is the real and imaginary axes and the portion of γR in the second quadrant. Now consider the complex integral: {z^{4}+1} &= {z^{4}+1} + {z^{4}+1} Now first look at the integral over γ₁. {z^{4}+1} &= {(z-({2}+i{2}))(z-(-{2}+i{2}))(z-(-{2}-i{2}))(z-({2}-i{2}))}\\ &= {(z-(-{2}+i{2}))(z-(-{2}-i{2}))(z-({2}-i{2}))}}{(z-({2}+i{2}))} Since the function in the numerator is holomorphic in γ₁, we can use Cauchy’s Integral formula. So we have, {(z-(-{2}+i{2}))(z-(-{2}-i{2}))(z-({2}-i{2}))}}{(z-({2}+i{2}))} &= {({2}+i{2}+{2}-i{2})({2}+i{2}+{2}+i{2})({2}+i{2}-{2}+i{2})}\\ &= {(\sqrt2)(\sqrt2+i\sqrt2)(i\sqrt2)}\\ &= {2i(\sqrt2 +i\sqrt2)}\\ &= {2i\sqrt2-2\sqrt2}\\ &={-\sqrt2+i\sqrt2}\\ And now for the integral over γ₂. {z^{4}+1} &= {(z-({2}+i{2}))(z-(-{2}+i{2}))(z-(-{2}-i{2}))(z-({2}-i{2}))}\\ &= {(z-({2}+i{2}))(z-(-{2}-i{2}))(z-({2}-i{2}))}}{(z-(-{2}+i{2}))} We may use Cauchy’s Integral Formula once again because the function in the numerator is holomorphic in γ₂. {(z-({2}+i{2}))(z-(-{2}-i{2}))(z-({2}-i{2}))}}{(z-(-{2}+i{2}))} &= {(-{2}+i{2}-{2}-i{2})(-{2}+i{2}+{2}+i{2})(-{2}+i{2}-{2}+i{2})}\\ &= {(-\sqrt2)(-\sqrt2+i\sqrt2)(i\sqrt2)}\\ &= {-2i(-\sqrt2 +i\sqrt2)}\\ &= {2i\sqrt2+2\sqrt2}\\ &={\sqrt2+i\sqrt2}\\ So we have, {z^{4}+1} &= {z^{4}+1} + {z^{4}+1}\\ &= {-\sqrt2+i\sqrt2} + {\sqrt2+i\sqrt2}\\ &= {(-\sqrt2 + i\sqrt2)(\sqrt2 + i\sqrt2)}\\ &= {-2-2}\\ &= {-4}\\ &= {2}\\ Now using Proposition 4.8(d) and the reverse triangle inequality, \left|{z^{4}+1}\right| &\leq max_{z\in\gamma_R}\left|{z^{4}+1}\right|\pi R\\ &\leq max_{z\in\gamma_R}\left({|z|^{4}-1}\right) = {R^{4}-1} Which goes to zero as R approaches infinity. Thus, {2} &= {z^{4}+1}\\ &= {z^{4}+1}+{z^{4}+1}\\ &={z^{4}+1}\\ &=^{\infty}{x^{4}+1}

Use the Cauchy Integral Formula(Theorem 4.30) to evaluate the integral in Exercise 4.33 when r=3. Consider the integral $$ {z^2-2z-8}dz= {(z+2)(z-4)}.$$ Notice the function $f(z)={z-4}$ is holomorphic in ℂ\{4} which contains $D\overline [-2,3]$. Thus, we can apply the Cauchy Integral Formula(Theorem 4.30): {2\pi i}{z-4}}{z+2}dz\\ =2\pi i f(-2)\\ =2\pi i (-{6})\\ = {3}\\ Hence, $$ {z^2-2z-8}={3}$$ By Theorem 4.23 (Cauchy’s Theorem): if f is holomorphic in ℂ \ {−2} and C[0, 3]∼ℂ \ {−2}C[−2, 3] then $${z^2-2z-8}= {z^2-2z-8} $$ Therefore,$${z^2-2z-8}={3}$$

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Fix a positive integer n and a complex number w. Find all solutions to zn = w. (Hint:write w in terms of polar coordinates.) Let n ∈ ℤ and w ∈ ℂ. So, in polar coordinates, w = reiθ for some r, θ ∈ ℝ. Notice that the r represents the modulus of the complex number and θ is the argument. So, we have zn = reiθ. Geometrically, we are multiplying the lengths and add their angles. We need to find values of z where it is multiplied by itself n times and is equal to reiθ. So, let z = seiθ. Thus, we have sn = r. Hence, $s=\sqrt[n]{r}.$ To solve for θ, we have nθ = θ + 2πk, for some k ∈ ℤ. Then $\theta={n}.$ Therefore we have $z=\sqrt[n]{r}e^{i{n}}.$ By the Fundamental Theorem of Algebra, there are n solutions to zn = w.

Fix a ∈ ℂ and b ∈ ℝ. Show that the equation |z²| + Re(az)+b = 0 has a solution if and only if |a²| ≥ 4b. When solutions exist, show the solution set is a circle. Let a ∈ ℂ, b ∈ ℝ and z ∈ ℂ. Then z = x + yi and a = c + di for some x, y, c, d ∈ ℝ. Then we have, $$\left|z^2\right|+Re(az)+b=0,$$ By substitution we get the following: $$x^2+y^2 + Re((c+di)(x+yi))+b =0$$ By algebra, $$x^2+y^2 + Re(cx+cyi+dxi+dyi^2)=-b$$ $$ x^2+y^2 + Re(cx-dy+i(dx+cy))=-b$$ $$ x^2 +y^2 + cx-dy = -b$$ $$(x^2+cx) +(y^2-dy)=-b$$ Then by completing the square, $$(x+{2}c)^2 + (y-{2}d)^2 = -b + {4}(c^2+d^2)$$ Thus, $$(x+{2}c)^2 + (y-{2}d)^2 = -b + {4}|a^2|$$ Notice the left hand side $$(x+{2}c)^2 + (y-{2}d)^2 \geq 0$$ Then the right hand side $$ -b + {4}|a^2|\geq 0.$$ Thus, ${4}|a^2|\geq b$ Hence, |a²|≥4b. Assume |a²|≥4b. Then ${4}|a^2|\geq b.$ Thus, ${4}(c^2+d^2)\geq b.$ Therefore, the equation |z²| + Re(az)+b = 0 has a solution if and only if |a²| ≥ 4b.

Consider the function f : ℂ/{0}→ℂ given by $f(x)={x}.$ Apply the definition of the derivative to give a direct proof that $f'(z)=-{z^2}.$ Let f : ℂ/{0}→ℂ be given by $f(x)={x}.$ Then by the definition of a derivative we have the following: f'(x) &= {h}\\ &={(z_0+h)}-{z_0}}{h}\\ &={z_0(z_0+h)}-{z_0(z_0+h)}}{h}\\ &={z_0(z_0+h)}}{h}\\ &={z_0(z_0+h)}}{h}\\ &= {z_0(z_0+h)}\\ &= {(z_0)^2+z_0h}\\ &= {(z_0)^2+z_0(0)}\\ Therefore, $f'(z) = {(z_0)^2}$, such that z₀ ≠ 0.

1.10 1.10 Fix a ∈ ℂ and b ∈ ℝ. Show that the equation |z²|+Re(az)+b = 0 has a solution if an only if |a²|≥4b. When solutions exist, show the solution set is a circle. Fix a ∈ ℂ and b ∈ ℝ. Suppose that the equation |z²|+Re(az)+b = 0 has a solution. Let z = x + iy and a = c + id So: |z^2|+Re(az)+b&=0\\ x^2+y^2+cx-dy+b&=0\\ x^2+cx+y^2-dy&=-b\\ x^2+cx+{4}c^2+y^2-dy+{4}d^2&=-b+{4}c^2+{4}d^2\\ (x+{2}c)^2+(y+{2}d)^2&=-b+{4}(c^2+d^2)\\ (x+{2}c)^2+(y+{2}d)^2&=-b+{4}|a^2| Thus the solution set is a circle centered at $-{2}+{2}i$ with radius ${4}|a^2|}$. By checking the left hand side, $(x+{2}c)^2+(y+{2}d)^2\geq0$. So the right hand side is: -b+{4}|a^2|&\geq0\\ {4}|a^2|&\geq b\\ |a^2|\geq4b Meanwhile, assume ∣a² ∣ ≥4b. So: \mid a^2 \mid &\geq 4b\\ a&\geq 4b &\\ c^2+d^2&\geq 4b\\ {4}&\geq b & \\ {4}-b &\geq 0\\ {4}+{4}-b &\geq 0 & Since the solution set is a circle, then |z²|+Re(az)+b = 0 has a solution when ∣a² ∣ ≥4b. Therefore |z²|+Re(az)+b = 0 has a solution if and only if |a²|≥4b, as was to be shown.

_Fix a positive integer_ n _and a complex number_ w. _Find all solutions to_ zn = w. Let n ∈ ℤ and w ∈ ℂ. So w = reiθ for some non-negative r ∈ ℝ and θ ∈ ℝ. Then, z^{n}&=re^{i\theta}\\ &=re^{i\theta}e^{(2\pi m)i}$}\\ &=re^{i(\theta+2\pi m)} Thus, z=\sqrt[n]{z^{n}}=\sqrt[n]{r}e^{{n}} If m > n, then by the division algorithm, m = qn + k for some q ∈ ℤ and some positive integer 0 ≤ k < n. So, e^{{n}} &=e^{{n}}e^{{n}}\\ &=e^{{n}}e^{{n}}\\ &=e^{{n}}e^{{n}}e^{{n}}\\ &=e^{{n}}e^{i2\pi q}e^{{n}}\\ &=e^{{n}}e^{{n}}\\ &=e^{{n}} Since 0 ≤ k < n, we need only to consider the solutions where 0 ≤ m < n. The final solution then is: $z=\sqrt[n]{z^{n}}=\sqrt[n]{r}e^{{n}}$ for all m ∈ ℤ such that 0 ≤ m < n.

_Fix an_ a ∈ ℂ _and_ b ∈ ℝ. _Show that the equation_ |z²|+Re(az)+b = 0 _has a solution if and only if_ |a²|≥4b. Let a ∈ ℂ,b ∈ ℝ, and z ∈ ℂ. Then a = d + ei and z = f + gi for some d, e, f, g ∈ ℝ. Note that: $$|z^{2}|=z\overline z=f^{2}+g^{2}$$ and $$Re(az)=fd-eg$$ So |z²|+Re(az)+b = 0 has solutions &\iff (f^{2}+g^{2})+(fd-eg)+b=0\\ &\iff f^{2}+fd+g^{2}-eg+b=0\\ &\iff (f+{2}d)^{2}-{4}d^{2}+(g-{2})^{2}-{4}e^{2}+b=0 \\ &\iff (f+{2}d)^{2}+(g-{2}e)^2={4}d^{2}+{4}e^{2}-b\\ &\iff(f+{2}d)^{2}+(g-{2}e)^2={4}(d^{2}+e^{2})-b\\ The above equation is a circle with radius $r={4}(d^{2}+e^{2})-b}$, so r² must be greater than or equal to zero in order for a solution to exist. Thus, {4}(d^{2}+e^{2})-b\geq0 &\iff(d^{2}+e^{2})\geq4b And since a = d + ei, $|a^{2}|=a\overline a=d^{2}+e^{2}$. Thus a solution to |z²|+Re(az)+b = 0 exists if and only if |a²|≥4b.