Calculate with complex numbers in Cartesian form

Extend the real numbers

A complex number is z=a+bi, with a,b real and i²=−1. Re(z)=a and Im(z)=b are real numbers. A real number has b=0; a non-real complex number has b≠0. Equality requires both corresponding components to agree

For z=3−4i, enter Im(z)

Powers of i repeat every four steps

The sequence 1,i,−1,−i repeats because i⁴=1. Reduce a nonnegative exponent modulo 4. Negative powers also follow from i⁻¹=−i.

Enter the imaginary part of i¹⁵

Add corresponding components

Add or subtract real parts and imaginary parts separately. Subtraction changes both signs of the second number. Argand addition is the same coordinate rule as plane-vector addition

Find the real part of (−2+3i)+(5−i)

Find the imaginary part of (−2+3i)+(5−i)

Multiply, then replace i²

Expand as for algebraic brackets, then use i²=−1. This gives (a+bi)(c+di)=(ac−bd)+(ad+bc)i.

Find the real part of (2+3i)(1−2i)

Find the imaginary part of (2+3i)(1−2i)

Conjugation reflects in the real axis

The conjugate of a+bi is a−bi. In an Argand diagram, use a horizontal real axis and vertical imaginary axis. Conjugation keeps the x-coordinate and reverses the y-coordinate. A real number equals its own conjugate

Which number is the conjugate of −2+3i?

Modulus is a distance

The modulus |a+bi|=√(a²+b²) is the distance from the origin. Hence z times its conjugate is |z|². Modulus is nonnegative and equals zero exactly when z=0.

Find |3−4i|

Use a conjugate to divide

For nonzero w=c+di, multiply numerator and denominator by c−di. The denominator becomes c²+d²>0. This gives the real and imaginary components separately. Division by zero is undefined

Find the real part when 3+4i is divided by 1−2i

Find the imaginary part when 3+4i is divided by 1−2i

Solve a complex linear equation

For wz+c=d with w≠0, subtract c and divide by w. Check both components by substitution. If w=0, the equation instead has every complex solution when c=d, and no solution otherwise

Solve (1+i)z+(2−i)=5+2i. Enter Re(z)

Use the discriminant over the stated domain

For Az²+Bz+C=0 with real A,B,C and A≠0, let D=B²−4AC. If D>0 there are two distinct real roots; if D=0 there is one real value with multiplicity two; if D<0 there are two distinct non-real conjugates. Use the quadratic formula in each case and state the solution domain

For 2z²−4z+5=0, find D.

Extract the imaginary factor carefully

When D<0, use the displayed quadratic formula. The symbol √(−D) is a positive real square root. Avoid applying √u√v=√(uv) indiscriminately to negative real inputs

For 2z²−4z+5=0, write the roots as h±k√d i with k>0. Enter h.

For 2z²−4z+5=0, write the roots as h±k√d i with k>0 and square-free d. Enter k.

For z²+9=0 over C, which set is complete?

A conjugate pair gives a real quadratic

For roots a±bi with b≠0, multiply their linear factors: (z−a−bi)(z−a+bi)=(z−a)²+b². For 2±3i the monic quadratic is z²−4z+13. A real-coefficient polynomial with one of these roots contains that factor, but can have additional factors. Its multiplicities are not determined by the given root alone

Justify conjugate pairing, ungraded

For a quadratic with real coefficients, prove that a non-real root has its conjugate as the other root. Then explain why the real-coefficient assumption is needed. This proof is ungraded

Model: conjugate the equation

For real A,B,C, conjugating Aα²+Bα+C=0 gives A(conjugate α)²+B(conjugate α)+C=0. Thus the conjugate is also a root. If Im(α)≠0, the roots are distinct. A real root is its own conjugate and need not be repeated. The polynomial z−(1+2i), whose constant is non-real, has 1+2i as its only root; its conjugate is not a root

Construct and explain an Argand sketch, ungraded

Plot 2+3i, its conjugate, its negative, and the sum of 2+3i and its conjugate. Label axes, equal unit scales and the origin. Explain the reflection and vector addition, and distinguish modulus from a signed component. This drawing and explanation are ungraded

Model: coordinates determine the construction

For 2+3i, its conjugate, its negative, and the sum of 2+3i and its conjugate, use (2,3), (2,−3), (−2,−3) and (4,0), respectively. The first two points reflect across the real axis and each has modulus √13. Their sum lies at (4,0). The negative lies opposite the original point through the origin. No root multiplicities follow from these coordinates alone

Use technology as a check, ungraded

Before running, select an application and functionality for complex arithmetic and exact equation solving. Justify the choice and explain one domain or precision limitation. In complex mode, evaluate the given product and quotient and solve the quadratic. Compare exact and decimal output, then substitute the roots into the original equation. Explain what a real-only solver may report. This technology task is ungraded. Retain the application/version, domain and exact/decimal settings, entered expressions and actual outputs. Supply the independent algebra or substitution check and your rationale for review

Model: preserve the domain and exact answer

A symbolic application with a complex-domain solver is a defensible choice because the task requires the exact non-real pair as well as decimal comparison. A real-only solver cannot supply that pair, and decimal-only output does not preserve its exact form. Multiplying 2+3i by 1−2i gives 8−i, and dividing 3+4i by 1−2i gives −1+2i. The roots of 2z²−4z+5=0 are 1±(√6/2)i, approximately 1±1.224745i. The displayed substitution gives zero exactly; the conjugate root also gives zero because the coefficients are real. A solver restricted to real roots may return no real solution; that does not mean no complex solution. Retain the exact pair as the answer and use decimals only as a check. Record your own application settings and outputs; these expected results are not evidence that you ran the tool