Mathematics

A. Algebra and trigonometry

Algebra and Trigonometry

The algebraic and trigonometric machinery the rest of the FE leans on: quadratics, logs, partial fractions, triangle laws, identities, and polar conversion.

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B. Complex numbers

Complex Numbers and Phasors

The three equivalent forms of a complex number, Euler's identity, complex arithmetic and roots, and the phasor representation that turns AC circuits into algebra.

Complex numbers are the single most leveraged idea on the FE-ECE: master them and AC circuits, impedance, transfer functions, and the Fourier and Laplace transforms all collapse from calculus into algebra. The whole trick is that a sinusoid

v(t)=V_m\cos(\omega t+\phi)

carries only two pieces of information at a fixed frequency, an amplitude and a phase, and a complex number stores exactly two numbers. The FE Reference Handbook's Algebra of Complex Numbers and Euler's Identity sections give you the forms and operations; this concept welds them to the phasor so you can read an AC problem and immediately write an impedance. Note the handbook uses

j

for

\sqrt{-1}

, since

i

denotes current.

The three forms

A complex number wears three interchangeable outfits. Rectangular form

z=a+jb

exposes the real part

a

and imaginary part

b

, and is the form for addition and subtraction. Polar form

z=c\angle\theta

gives magnitude

c

and angle

\theta

, and is the form for multiplication, division, and powers. Exponential form

z=ce^{j\theta}

is polar written so the algebra of exponents does the work for you. They describe the same point in the complex plane; fluency is the ability to switch on demand.

z=a+jb=c\angle\theta=ce^{j\theta},\qquad c=\sqrt{a^2+b^2},\;\; \theta=\arctan\!\tfrac{b}{a}

Euler's identity

Euler's identity is the hinge between the trigonometric and exponential worlds and the reason phasors work at all. From

e^{j\theta}=\cos\theta+j\sin\theta

you recover the real and imaginary parts of any unit phasor, and adding the conjugate relation

e^{-j\theta}=\cos\theta-j\sin\theta

lets you write

\cos\theta

and

\sin\theta

themselves as sums of exponentials. This is exactly how a cosine becomes the real part of a rotating phasor

\operatorname{Re}\{e^{j\omega t}\}

.

e^{j\theta}=\cos\theta+j\sin\theta,\qquad \cos\theta=\tfrac{e^{j\theta}+e^{-j\theta}}{2},\;\; \sin\theta=\tfrac{e^{j\theta}-e^{-j\theta}}{2j}

Arithmetic: add in rectangular, multiply in polar

Pick the form that makes the operation trivial. Add and subtract in rectangular form, combining real with real and imaginary with imaginary. Multiply and divide in polar form, where magnitudes multiply or divide and angles add or subtract. Trying to multiply two numbers in rectangular form is legal but slow and error-prone; trying to add in polar form is worse. The conjugate

z^{*}=a-jb

reflects across the real axis, and

z\,z^{*}=a^2+b^2=c^2

is the squared magnitude, the move that rationalizes a division.

z_1 z_2=c_1 c_2\,\angle(\theta_1+\theta_2),\qquad \dfrac{z_1}{z_2}=\dfrac{c_1}{c_2}\,\angle(\theta_1-\theta_2)

Powers and roots

Raising to a power is effortless in polar form by De Moivre's theorem: magnitude to the power, angle times the power. Roots run the same machine in reverse, and here the FE likes to catch you, because every nonzero complex number has exactly

k

distinct

k

-th roots spaced evenly by

360^\circ/k

around a circle of radius

c^{1/k}

. You get them by adding

360^\circ n

to the angle before dividing, for

n=0,1,\dots,k-1

.

z^{1/k}=c^{1/k}\,\angle\dfrac{\theta+360^\circ n}{k},\qquad n=0,1,\dots,k-1

From sinusoid to phasor

A phasor is the complex amplitude of a sinusoid at a known frequency: strip off the

\cos(\omega t)

carrier and keep the magnitude and phase. The universal convention on the FE is the cosine reference, so

v(t)=V_m\cos(\omega t+\phi)

maps to the phasor

\mathbf{V}=V_m\angle\phi

(or

V_{\text{rms}}\angle\phi

when working in RMS). A sine must first be converted with

\sin\theta=\cos(\theta-90^\circ)

. The frequency

\omega

does not appear in the phasor; it returns only when you map back to the time domain.

v(t)=V_m\cos(\omega t+\phi)\;\longleftrightarrow\;\mathbf{V}=V_m\angle\phi

Impedance: why phasors earn their keep

Once voltages and currents are phasors, the calculus of capacitors and inductors becomes multiplication by a complex impedance. A resistor is

Z_R=R

, an inductor

Z_L=j\omega L

, and a capacitor

Z_C=1/(j\omega C)=-j/(\omega C)

. Ohm's law then reads

\mathbf{V}=\mathbf{I}\,Z

in the complex plane, so series and parallel combinations, dividers, and Thevenin reductions all work exactly as in DC, just with complex numbers. The angle of

Z

is the phase by which voltage leads current.

Z_R=R,\qquad Z_L=j\omega L,\qquad Z_C=\dfrac{1}{j\omega C}=-\dfrac{j}{\omega C}

Exam strategy

Decide the form before you compute: rectangular for sums, polar for products, quotients, powers, and roots, and convert once rather than fighting the wrong form. Keep angle mode and the

j

-versus-

i

convention straight. For phasors, convert every source to the cosine reference up front and carry RMS or peak consistently through the whole problem. When a question asks for all roots, remember there are exactly

k

of them spaced

360^\circ/k

apart, and the most common wrong answer reports only one.

Key equations

Three equivalent forms

z=a+jb=c\,\angle\theta=c\,e^{j\theta}

Rectangular ( $a$ real, $b$ imag), polar ( $c$ magnitude, $\theta$ angle), exponential. Same point in the complex plane.

Magnitude and angle

c=\sqrt{a^2+b^2},\qquad \theta=\arctan\!\big(\tfrac{b}{a}\big)\;(\text{quadrant-corrected})

Rectangular to polar. Quadrant-correct $\theta$ from the signs of $a$ and $b$ (or use $\operatorname{atan2}(b,a)$ ).

Rectangular components

a=c\cos\theta,\qquad b=c\sin\theta

Polar to rectangular: real and imaginary parts of $c\angle\theta$ .

Euler's identity

e^{\pm j\theta}=\cos\theta\pm j\sin\theta

Links exponential and trigonometric forms; the basis of phasors and Fourier analysis.

Addition / subtraction

z_1\pm z_2=(a_1\pm a_2)+j(b_1\pm b_2)

Use rectangular form. Combine real with real, imaginary with imaginary.

Multiplication / division

z_1 z_2=c_1 c_2\,\angle(\theta_1+\theta_2),\qquad \tfrac{z_1}{z_2}=\tfrac{c_1}{c_2}\,\angle(\theta_1-\theta_2)

Use polar form. Multiply/divide magnitudes, add/subtract angles.

Conjugate and magnitude product

z^{*}=a-jb,\qquad z\,z^{*}=a^2+b^2=c^2

Conjugate reflects across the real axis; product gives squared magnitude. Used to rationalize division.

De Moivre (powers)

z^{n}=c^{n}\,\angle(n\theta)=c^{n}(\cos n\theta+j\sin n\theta)

Raise to integer power $n$ in polar form: magnitude to the power, angle times the power.

k-th roots

z^{1/k}=c^{1/k}\,\angle\dfrac{\theta+360^\circ n}{k},\;\; n=0,\dots,k-1

Every nonzero $z$ has exactly $k$ distinct roots, evenly spaced by $360^\circ/k$ .

Sinusoid to phasor (cosine reference)

V_m\cos(\omega t+\phi)\;\leftrightarrow\;\mathbf{V}=V_m\angle\phi

Phasor = complex amplitude. Convert sines via $\sin\theta=\cos(\theta-90^\circ)$ first.

Element impedances

Z_R=R,\;\; Z_L=j\omega L,\;\; Z_C=\dfrac{1}{j\omega C}

Phasor-domain impedances; $\omega=2\pi f$ in rad/s, $R$ in $\Omega$ , $L$ in H, $C$ in F.

Phasor Ohm's law

\mathbf{V}=\mathbf{I}\,Z

Complex Ohm's law. $\angle Z$ is the phase by which voltage leads current.

Worked examples

Convert, multiply, and divide

Problem. Given $z_1=3+j4$ and $z_2=1-j2$ , find $z_1$ and $z_2$ in polar form, then compute $z_1 z_2$ and $z_1/z_2$ .

Solution. Polar:

z_1=\sqrt{3^2+4^2}\angle\arctan(4/3)=5\angle53.1^\circ

;

z_2=\sqrt{1^2+2^2}\angle\arctan(-2/1)=2.24\angle{-63.4^\circ}

. Product (add angles):

z_1 z_2=(5)(2.24)\angle(53.1^\circ+(-63.4^\circ))=11.2\angle{-10.3^\circ}

. Check in rectangular:

(3+j4)(1-j2)=3-j6+j4+8=11-j2

, whose magnitude

\sqrt{121+4}=11.2

and angle

\arctan(-2/11)=-10.3^\circ

agree. Quotient (subtract angles):

z_1/z_2=\tfrac{5}{2.24}\angle(53.1^\circ-(-63.4^\circ))=2.24\angle116.6^\circ

. Note the magnitude is

5/\sqrt{5}=\sqrt{5}=2.24

, which only coincidentally equals

|z_2|

. Final:

z_1 z_2=11.2\angle{-10.3^\circ}

,

z_1/z_2=2.24\angle116.6^\circ

.

z_1 z_2=11.2\angle{-10.3^\circ},\qquad \tfrac{z_1}{z_2}=2.24\angle116.6^\circ

Cube roots of a complex number

Problem. Find all cube roots of $z=8\angle60^\circ$ .

Solution. Magnitude of each root:

8^{1/3}=2.00

. Angles:

\dfrac{60^\circ+360^\circ n}{3}

for

n=0,1,2

, giving

20^\circ

,

140^\circ

, and

260^\circ

. So the three roots are

2\angle20^\circ

,

2\angle140^\circ

,

2\angle260^\circ

. Sanity check: the angles are spaced exactly

120^\circ=360^\circ/3

apart, as a complete set of cube roots must be. Cubing the first:

2^3\angle(3\cdot20^\circ)=8\angle60^\circ

, recovering

z

. Final:

2\angle20^\circ,\;2\angle140^\circ,\;2\angle260^\circ

.

z^{1/3}=2\,\angle\{20^\circ,\,140^\circ,\,260^\circ\}

Sinusoid to phasor and impedance

Problem. A source $v(t)=170\cos(377t+30^\circ)\,\text{V}$ drives a series $R=10\,\Omega$ and $L=20\,\text{mH}$ . Find the phasor voltage (peak), the impedance, and the current magnitude.

Solution. Phasor (peak):

\mathbf{V}=170\angle30^\circ\,\text{V}

. Inductive reactance:

X_L=\omega L=377(0.020)=7.54\,\Omega

, so

Z=10+j7.54=12.5\angle37.0^\circ\,\Omega

. Current:

\mathbf{I}=\mathbf{V}/Z=\dfrac{170\angle30^\circ}{12.5\angle37.0^\circ}=13.6\angle{-7.0^\circ}\,\text{A}

, so

|I|=13.6\,\text{A}

peak. Sanity check: the current lags the voltage by

37^\circ

(the impedance angle), correct for an inductive load. Final:

\mathbf{V}=170\angle30^\circ\,\text{V}

,

Z=12.5\angle37.0^\circ\,\Omega

,

|I|=13.6\,\text{A}

peak.

Z=10+j7.54=12.5\angle37.0^\circ\,\Omega

Common pitfalls

•Adding or subtracting in polar form, or multiplying/dividing in rectangular form: pick the form that makes the operation a one-liner.
•Reporting only one root when $k$ are asked for. A $k$ -th root has exactly $k$ values spaced $360^\circ/k$ apart.
•Forgetting to convert a sine source to the cosine reference ( $\sin\theta=\cos(\theta-90^\circ)$ ) before reading off a phasor angle.
•Mixing RMS and peak in the same problem; the phasor magnitude must be one or the other consistently throughout.
•Sign slip on capacitive impedance: $Z_C=1/(j\omega C)=-j/(\omega C)$ is negative imaginary, so current leads voltage.
•Quadrant error: $\arctan(b/a)$ from a calculator can land in the wrong half-plane; check the signs of $a$ and $b$ .
•Treating $\omega$ as part of the phasor; the phasor holds only magnitude and phase, and $\omega$ re-enters only on the return to time domain.

References

NCEES FE Reference Handbook — Mathematics: Algebra of Complex Numbers and Polar Coordinate System
NCEES FE Reference Handbook — Mathematics: Euler's Identity and Roots
NCEES FE Reference Handbook — Electrical and Computer Engineering: AC Circuits

C. Discrete mathematics

Discrete Mathematics

Sets, relations and functions, propositional logic and truth tables, graphs and digraphs, counting, and recurrences at the depth the FE-ECE actually tests.

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D. Analytic geometry

Analytic Geometry

Lines and distance, the four conic sections and how to classify them, and conversion among polar, cylindrical, and spherical coordinate systems.

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E. Calculus

Calculus

Limits, derivatives (including partials), optimization, integrals, the fundamental theorem, and Taylor/L'Hopital tools the FE math block leans on every form.

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F. Ordinary differential equations

Ordinary Differential Equations

Solve first-order linear/separable ODEs and second-order constant-coefficient equations, classify over/critical/under-damped responses, and fit initial conditions.

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G. Linear algebra

Linear Algebra

Matrix algebra, determinants, inverses, Cramer's rule and Gaussian elimination, rank and singularity, and the eigenvalue/eigenvector problem behind FE systems work.

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H. Vector analysis

Vector Analysis

Dot, cross, and triple products; gradient, divergence, and curl; line and surface integrals; plus an intro to the divergence and Stokes theorems used across fields.

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