求一个复数 a + bi 使得 a 2 + b 2是无理数
实数和虚数结合形成复数。虚部 I (iota) 表示 -1 的平方根。复数的虚部是 i。 a + ib 是矩形或标准形式的复数的典型表示。例如,420 + 69i 是一个复数,其中 420 代表实部,69 代表虚部。
模数
当一个复数出现在图表上时,它的实部绘制在 x 轴上,虚部绘制在 y 轴上。假设如果数字由下图中的点 P 表示,三角形 OPA 和 OPB 都是直角的。显然,在直角三角形 POA 中,PO 是斜边; Oa 是底,Pa 是垂线。使用毕达哥拉斯定理,我们有:
OP 2 = OA 2 + PA 2
OP =
复数的绝对值被视为其模数。它是实部和虚部平方和的平方根。在上述情况下,OP 是 z = a + ib 形式的复数的模,用 r 表示。
求一个复数 a + bi 使得 a 2 + b 2是无理数。
解决方案:
An irrational number is the one which can be expressed in the form of a/b, where b ≠ 0, like √2, √3, etc.
Say we are given the complex number 1 + .
Comparing this with the form a + ib, we have:
a = 1, b =
Now, a2 + b2 =
= 1 + 42/3
Clearly, 42/3 cannot be expressed in the form of a/b. Hence it is an irrational number.
类似问题
问题 1. 证明模数的平方是不合理的。
解决方案:
Comparing this with the form a + ib, we have:
a = 0, b =
Now, a2 + b2 =
= 22/3
Clearly, 22/3 cannot be expressed in the form of a/b. Hence it is an irrational number.
问题 2. 证明模数的平方是不合理的。
解决方案:
Comparing this with the form a + ib, we have:
a = 0, b =
Now, a2 + b2 =
= 22/5
Clearly, 22/5 cannot be expressed in the form of a/b. Hence it is an irrational number.
问题 3. 证明 3 + 的模的平方是不合理的。
解决方案:
Comparing this with the form a + ib, we have:
a = 3, b =
Now, a2 + b2 =
= 9 + 22/3
Clearly, 22/3 cannot be expressed in the form of a/b. Hence it is an irrational number.
问题 4. 证明模数的平方是理性的。
解决方案:
Comparing this with the form a + ib, we have:
a = 4, b = = 2
Now, a2 + b2 = 42 + 22
= 20
Clearly, 20 can be expressed in the form of a/b. Hence it is a rational number.
问题 5. 证明 10 + 模的平方是不合理的。
解决方案:
Comparing this with the form a + ib, we have:
a = 10, b =
Now, a2 + b2 =
= 100 + 22/3
Clearly, 22/3 cannot be expressed in the form of a/b. Hence it is an irrational number.