Complex Numbers

1. \(z = 4 + 3i\): \(\operatorname{Re}(z) = 4\), \(\operatorname{Im}(z) = 3\).

2. \(z = -5i\): \(\operatorname{Re}(z) = 0\), \(\operatorname{Im}(z) = -5\) (pure imaginary).

3. \(z = 7\): \(\operatorname{Re}(z) = 7\), \(\operatorname{Im}(z) = 0\) (pure real).

4. \(z = -2 + i\sqrt{5}\): \(\operatorname{Re}(z) = -2\), \(\operatorname{Im}(z) = \sqrt{5}\).

Express \(\sqrt{-50}\) in the form \(x + iy\).

Solution.

Solve \(z^2 + z + 1 = 0\).

Solution.

Thus, the solutions are

Represent \(1 + i\sqrt{3}\) in polar form.

Solution.

Thus:

Convert \(3 e^{-i\pi/2}\) to rectangular form.

Solution.

So:

Write \(z=-1-i\) in polar form.

Solution.

Hence

Let \( z=\dfrac{2-3i}{1+i} \). Compute \( |z| \), \( \operatorname{Re}(z) \), and \( \operatorname{Im}(z) \).

Solution.

Thus \( \operatorname{Re}(z)=-\tfrac{1}{2},\ \operatorname{Im}(z)=-\tfrac{5}{2} \), and

(Alternatively, \( |z|=\sqrt{z\overline{z}} \).)

\( \dfrac{2+i}{3-i} = \dfrac{(2+i)(3+i)}{(3-i)(3+i)} = \dfrac{6+5i+i^2}{3^2-(-1)} = \dfrac{5+5i}{10} = \frac12+\frac12 i. \)

\( \displaystyle \left|\frac{\sqrt{5}+3i}{\,1-i\,}\right| =\frac{\sqrt{(\sqrt{5})^2+3^2}}{\sqrt{1^2+(-1)^2}} =\frac{\sqrt{14}}{\sqrt{2}}=\sqrt{7}. \)

Solve \((x+iy)^2=2i\) for real \(x,y\).

Solution.

Equate real and imaginary parts:

From \(x^2-y^2=(x-y)(x+y)=0\) we have \(y=x\) or \(y=-x\).

Case 1: \(y=x\). Then \(2xy=2\Rightarrow 2x^2=2\Rightarrow x^2=1\). Hence \(x=\pm 1\) and \(y=x\). This gives \(1+i\) and \(-1-i\).

Case 2: \(y=-x\). Then \(2xy=2\Rightarrow 2x(-x)=2\Rightarrow -2x^2=2\Rightarrow x^2=-1\), which is impossible for real \(x\). So no solutions from this branch.

Therefore the solutions are

i.e. \(z=1+i\) and \(z=-1-i\).

Check. \((1+i)^2=1+2i-1=2i\) and \((-1-i)^2=1+2i-1=2i\).

(Alternative via polar form.) \(2i=2e^{i\pi/2}\). Square roots have modulus \(\sqrt{2}\) and arguments \(\frac{\pi}{4}\) and \(\frac{\pi}{4}+\pi\), giving \(\sqrt{2}e^{i\pi/4}=1+i\) and \(\sqrt{2}e^{i5\pi/4}=-1-i\).

1. \(|z-1|=2\) is the set of points at distance \(2\) from \(1\), i.e. the circle \((x-1)^2+y^2=4\).

2. \(\arg z = \pi/4\) is the ray \(y=x\) with \(x>0\) — a ray, not the whole line, because the argument of a point on the opposite side is \(\pi/4 - \pi = -3\pi/4\).

3. \(|z-1| < |z+1|\) is the set of points closer to \(1\) than to \(-1\), which is the half-plane \(x > 0\).

Test

Here

Thus the series converges absolutely (and hence converges).

Test

The terms cycle through \(i,-1,-i,1,\dots\). Real part: \(\sum \frac{\pm 1}{\sqrt{n}}\) alternating — converges by the alternating series test. Imaginary part: \(\sum \frac{\pm 1}{\sqrt{n}}\) alternating — also converges. Hence the series converges (conditionally, not absolutely).

converges iff \(|r|<1\), in which case

Ratio test:

Thus \(R = 1\), so the disk of convergence is \(|z|<1\).

Ratio test:

Thus convergence requires \(|z+1-i| < 3\). The disk is centered at \(z_0 = -1 + i\) with radius \(R = 3\).

Two sources produce oscillations of the same frequency \(\omega\) but different amplitudes and phases,

Find their sum. This is the central calculation in interference and in AC circuit analysis.

Solution.

Done with trigonometric identities this is tedious. The complex method is to notice that, by Euler's formula, \(A\cos(\omega t+\phi)\) is the real part of \(A e^{i(\omega t + \phi)}\). So write each wave as the real part of a complex exponential and add:

The time dependence \(e^{i\omega t}\) factors out completely, and everything specific to the two waves is collected in the bracket

called the phasor of the combined wave. Adding two oscillations has been reduced to adding two complex numbers — that is, to placing two arrows tip to tail in the complex plane. Reading off the modulus and argument of the sum,

and the answer is \(y_1 + y_2 = A\cos(\omega t + \phi)\): the sum is another oscillation at the same frequency, with a new amplitude and phase.

Two special cases are worth noting. If \(\phi_1 = \phi_2\) the phasors are parallel and \(A = A_1 + A_2\) — constructive interference. If \(\phi_2 = \phi_1 + \pi\) they are antiparallel and \(A = |A_1 - A_2|\) — destructive interference, which vanishes entirely when the amplitudes are equal.

Write \(1+i\) in polar form.

Solution.

The point \(1+i\) sits in the first quadrant, so \(\arctan\) needs no quadrant correction:

Thus

Polar form is what makes powers and roots easy; we use this result in Section 2.8.

Compute \((1+i)^8\).

Solution.

Using the polar form just found, raising to a power multiplies the angle by the exponent and raises the modulus to it:

The angle \(8\pi/4 = 2\pi\) is a full turn, which returns to the positive real axis — so the answer is real. Doing this by repeated multiplication in rectangular form would take eight expansions.

Find the cube roots of \(8e^{i\pi}\).

Solution.

Here \(r=8\) and \(\theta=\pi\). Taking the cube root divides the angle by \(3\), but the angle is only defined up to \(2\pi\), so all three choices must be included:

Writing out the three values,

They all have modulus \(2\) and are spaced \(2\pi/3\) apart, so they sit at the vertices of an equilateral triangle on the circle of radius \(2\).

Take for example \( f(z)=\dfrac{z^2+1}{z-3} \). Find \( f(i-2) \).

Multiply numerator and denominator by the conjugate \(-i-5\):

Expand the numerator and simplify:

Denominator:

Therefore

Compute \( e^{2+i}\,e^{-1+3i} \) without expanding series.

Solution.

Use property (a) to add exponents first, then property (c) to separate real and imaginary parts:

Note the division of labour in \(e^{x+iy} = e^x(\cos y + i \sin y)\): the real part of the exponent sets the size \(e^x\), and the imaginary part sets the angle \(y\). This one fact drives the rest of the chapter.

A resistor \(R\), an inductor \(L\) and a capacitor \(C\) are connected in series across a voltage source \(V(t) = V_0\cos(\omega t)\). Find the amplitude of the resulting current.

Solution.

The physical difficulty is that the three components respond differently: the voltage across a resistor is in phase with the current, across an inductor it leads by \(\pi/2\), and across a capacitor it lags by \(\pi/2\). Combining three oscillations with different phase shifts is exactly the problem of the previous phasor example.

Represent the voltage as \(\operatorname{Re}\big[V_0 e^{i\omega t}\big]\) and look for a current \(\operatorname{Re}\big[I_0 e^{i\omega t}\big]\). Because \(\frac{d}{dt}e^{i\omega t} = i\omega e^{i\omega t}\), differentiation becomes multiplication by \(i\omega\), and each component contributes a complex impedance \(Z\) with \(V = ZI\):

The factor \(i\) is doing the physics: multiplying by \(i = e^{i\pi/2}\) is a rotation by a quarter turn, which is precisely the \(\pi/2\) phase shift. In series the impedances add just as resistances do:

The current amplitude is then \(I_0 = V_0/|Z|\), so

The current is largest when the imaginary part of \(Z\) vanishes, at \(\omega = 1/\sqrt{LC}\) — this is resonance, and it is how a radio selects one station out of many. A differential equation has been replaced by the arithmetic of complex numbers.

Compute \((1+i)^8\).

Write

Then

Compute \((-\sqrt{3}+i)^3\).

Here

Thus

Fourth roots of \(-16\).

We write

With \(n=4\), the roots are

Explicitly:

Cube roots of \(8e^{i\pi}\).

Here \(r = 8\), \(\theta = \pi\), \(n=3\). Thus

i.e.

Solve \(w^5=1\).

We write

Then the solutions are

These five points form a regular pentagon on the unit circle.

1. \(\displaystyle \cos(2i)=\frac{e^{i(2i)}+e^{-i(2i)}}{2} =\frac{e^{-2}+e^{2}}{2}\equiv \cosh(2)=3.762\ldots\)

2. \(\displaystyle \sin\!\Big(\frac{\pi}{2}+i\ln(2)\Big)\)

   \begin{equation} \begin{aligned} \sin\!\Big(\tfrac{\pi}{2}+i\ln(2)\Big) &=\frac{e^{\,i(\pi/2+i\ln(2))}-e^{-\,i(\pi/2+i\ln(2))}}{2i}\\[2mm] &=\frac{e^{\,i\pi/2-\ln(2)}-e^{-\,i\pi/2+\ln(2)}}{2i}\\[2mm] &=\frac{i\,e^{-\ln|2|}+\,i\,e^{\ln(2)}}{2i} \qquad(\text{since }e^{i\pi/2}=i,\ e^{-i\pi/2}=-i)\\[2mm] &=\frac{i\cdot\frac{1}{2}+2i}{2i} =\Big(\frac{1}{2}+2\Big)\frac{1}{2} =\frac{1}{4}+1=\boxed{\frac{5}{4}}. \end{aligned} \end{equation}

Prove that \(\sin^{2}(z)+\cos^{2}(z)=1\) for every complex \(z\).

Solution.

Adding the two, the \(e^{2iz}\) and \(e^{-2iz}\) terms cancel and only the constants survive:

Note this is now an algebraic fact about exponentials, with no triangles anywhere. It holds for all complex \(z\), even where \(|\sin z|\) exceeds \(1\).

Show that \(\dfrac{d}{dz}\cos(z)=-\sin(z)\).

Solution.

Show that \(\cosh^2(z)-\sinh^2(z)=1\)

Hence

1. Find all values of \(\ln(-1)\).

   Solution.

   Write \(-1\) in polar form. Its modulus is \(1\) and its argument is \(\pi\), but any odd multiple of \(\pi\) names the same point:

   \begin{equation} -1 = e^{i(2n+1)\pi}, \qquad n \in \mathbb{Z}. \end{equation}

   Therefore

   \begin{equation} \ln(-1) = \ln(1) + i(2n+1)\pi = i(2n+1)\pi, \qquad n \in \mathbb{Z}, \end{equation}

   that is

   \begin{equation} \ln(-1) = \ldots,\ -i3\pi,\ -i\pi,\ i\pi,\ i3\pi,\ \ldots \quad\text{and}\quad \mathrm{Ln}(-1) = i\pi , \end{equation}

   the principal value being the one with \(\Theta = \pi \in (-\pi, \pi]\), which is the case \(n = 0\). Note that \(\ln(-1)\) has no real values at all: the logarithm of a negative number does not exist in \(\mathbb{R}\), but it exists, infinitely many times over, in \(\mathbb{C}\).

2. Find all values of \(\ln(1+i)\).

   Solution.

   Here \(|1+i| = \sqrt2\) and \(\arg(1+i) = \pi/4\), so

   \begin{equation} \ln(1+i)=\ln\sqrt{2}+i\bigg(\frac{\pi}{4}+2n\pi\bigg),\qquad n\in\mathbb{Z}, \end{equation}

   with principal value \(\mathrm{Ln}(1+i) = \tfrac12\ln 2 + i\pi/4\).

1. Find all the values of \(i^{-2i}\):

   \begin{equation} i^{-2i}=e^{-2i\ln(i)} =e^{-2i\,[\,\ln(1)+(\tfrac{\pi}{2}+2\pi n)i\,]} =e^{\,2(\tfrac{\pi}{2}+2\pi n)} =e^{\,\pi+4\pi n},\qquad n\in\mathbb{Z}. \end{equation}

2. \(i^{1/2}\):

   \begin{equation} i^{1/2}=e^{(1/2)\ln(i)} =e^{\frac{1}{2}\,[\,\ln(1)+(\tfrac{\pi}{2}+2\pi n)i\,]} =e^{\,i(\tfrac{\pi}{4}+\pi n)} =e^{\,i\pi/4}\,e^{\,i\pi n} =\pm\,\frac{1+i}{\sqrt{2}}. \end{equation}

Solve \(z=\cos^{-1}(2)\).

Solution.

For a real variable this has no solution, since \(\cos x\) never exceeds \(1\). Over \(\mathbb{C}\) it does. Start from the definition of \(\cos z\) and put \(U = e^{iz}\):

Multiplying through by \(U\) turns this into a quadratic:

So \(e^{iz}=2\pm\sqrt{3}\), a positive real number. Taking the complex logarithm, and remembering it is multivalued,

Dividing by \(i\) (that is, multiplying by \(-i\)),

Finally, note that \((2+\sqrt3)(2-\sqrt3) = 4 - 3 = 1\), so the two numbers are reciprocals and \(\ln(2-\sqrt{3})=-\ln(2+\sqrt{3})\). The two sign choices therefore collapse into a single \(\pm\):

The answer is genuinely complex, with no real part other than \(2n\pi\): asking for an angle whose cosine is \(2\) forces us off the real axis.

Show that \(\sinh^{-1}(x/a) = \ln\big(x+\sqrt{x^2+a^2}\big) - \ln a\).

Solution.

Let \(z=\sinh^{-1}\!\big(\frac{x}{a}\big)\). Then

Only the \(+\) sign is admissible: with the \(-\) sign the bracket is negative, and a real logarithm would not exist. Absorbing the constant \(-\ln a\) into the constant of integration gives the table entry: