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Suppose A = B and both are complex.
How many real facts do we get? Two.
Let us see why that is so useful.
a + bi = c + di if and only if a = c and b = d.
So one complex equation gives two real equations.
Book Example 4 ends with x = 3 and y = 2.
We want p + iq with (p + iq)2 = x + iy.
Comparing parts gives x = p2 − q2 and y = 2pq.
The book's formula (v) uses the modulus |z| = √(x2 + y2).
Book Example 5: 5 + 12i. Here |z| = 13.
p = √((13 + 5)/2) = 3 and q = √((13 − 5)/2) = 2.
The square roots are 3 + 2i and −3 − 2i.
A complex polynomial has complex coefficients. Example: (1 − i)z + 3i.
Fundamental theorem of algebra (book): a polynomial of degree n ≥ 1 has exactly n roots in ℂ.
So P(z) = a(z − z1)(z − z2)…(z − zn).
Example: z2 + 4 = (z + 2i)(z − 2i). No real factors exist, but complex ones do.
Book Example 6: z2 + (i − 3)z − 3i = (z + i)(z − 3).
Book Example 9: z3 − 3z2 + z + 5 = (z + 1)(z − 2 − i)(z − 2 + i).
Book Example 10 (completing the square): 2z2 − 12z + 50 = 0 gives z = 3 ± 4i.
Step 1 / 7
Suppose A = B and both are complex.
How many real facts do we get? Two.
Let us see why that is so useful.
a + bi = c + di if and only if a = c and b = d.
So one complex equation gives two real equations.
Book Example 4 ends with x = 3 and y = 2.
We want p + iq with (p + iq)2 = x + iy.
Comparing parts gives x = p2 − q2 and y = 2pq.
The book's formula (v) uses the modulus |z| = √(x2 + y2).
Book Example 5: 5 + 12i. Here |z| = 13.
p = √((13 + 5)/2) = 3 and q = √((13 − 5)/2) = 2.
The square roots are 3 + 2i and −3 − 2i.
A complex polynomial has complex coefficients. Example: (1 − i)z + 3i.
Fundamental theorem of algebra (book): a polynomial of degree n ≥ 1 has exactly n roots in ℂ.
So P(z) = a(z − z1)(z − z2)…(z − zn).
Example: z2 + 4 = (z + 2i)(z − 2i). No real factors exist, but complex ones do.
Book Example 6: z2 + (i − 3)z − 3i = (z + i)(z − 3).
Book Example 9: z3 − 3z2 + z + 5 = (z + 1)(z − 2 − i)(z − 2 + i).
Book Example 10 (completing the square): 2z2 − 12z + 50 = 0 gives z = 3 ± 4i.
Q1. Find the square root of 5 + 12i.
Here x = 5, y = 12, so |z| = 13. √((13 + 5)/2) = 3 and √((13 − 5)/2) = 2. The roots are 3 + 2i and −3 − 2i. Check: (3 + 2i)2 = 5 + 12i.
Q2. Factorise z2 + 4 over ℂ.
z2 + 4 = 0 gives z2 = −4, so z = ±2i. Then z2 + 4 = (z + 2i)(z − 2i). Check: (z + 2i)(z − 2i) = z2 − 4i2 = z2 + 4.
✗ “The square root of 5 + 12i is only 3 + 2i.”
✓ There are two: 3 + 2i and −3 − 2i.
✗ “From (3x − 2y) + (2x + 3y)i = 5 + 12i we get 3x − 2y + 2x + 3y = 17.”
✓ Do not mix real and imaginary parts. Write 3x − 2y = 5 and 2x + 3y = 12 separately.
1. What is the square root of 5 + 12i that has a positive real part?
(a) (3 + 2i)2 = 9 + 12i − 4 = 5 + 12i.
2. What is |5 + 12i|?
(b) √(52 + 122) = √169 = 13.
3. If a + bi = c + di, then:
(a) Book section 1.2: equal when real parts and imaginary parts are equal.
4. How many roots does a polynomial of degree 5 have in ℂ?
(c) The book: degree n ≥ 1 means exactly n roots in ℂ.
5. Factorise z2 + 4 over ℂ.
(b) The book: z2 + 4 = (z + 2i)(z − 2i).
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