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Last modified on February 22nd, 2024
For a complex number z = a + ib, the absolute or modulus value measures the linear distance from the origin (0, 0) to the point (a, b). But unlike the absolute value of real numbers, it is defined in the argand (complex) plane, as shown.
In the argand plane, the x-axis represents the real part, while the y-axis represents the complex number’s imaginary part.
If (a, b) is represented by the point P, then to find the absolute value |z|, we will join the point P with the origin O. Similarly, by joining P with A (a, 0), we get a right-angled triangle OPA at A. where the horizontal distance of P is ‘a’, and the vertical distance of P is ‘b.’
Now, applying the Pythagoras theorem to the triangle OPA, we get OP = |z| = ${\sqrt{a^{2}+b^{2}}}$, which represents the hypotenuse of the right-angled triangle.
Thus, the absolute value of z is given by the formula:
|z| = ${\sqrt{a^{2}+b^{2}}}$
Now, let us find the absolute value of a complex number z = 6 + 8i is ${\sqrt{6^{2}+8^{2}}}$ = ${\sqrt{100}}$ = 10.
Complex numbers can have an absolute value of 1. It is the same for -1, just as for the imaginary numbers i and -i. This is because all of them are one unit away from 0, either on the real number line or the imaginary axis. It includes all the complex numbers of absolute value 1. Thus, the equation of the unit circle is |z| = 1.
If two complex numbers are z = 5 + 4i and -z = -5 – 4i.
Then, we have |z| = ${\sqrt{5^{2}+4^{2}}}$ and
|-z| = ${\sqrt{\left( -5\right) ^{2}+\left( -4\right) ^{2}}}$ = ${\sqrt{5^{2}+4^{2}}}$
For example, if z = 5 + 2i, then, |z2| = |(5 + 2i)(5 + 2i)| = |21 + 20i|
= ${\sqrt{21^{2}+20^{2}}}$ = ${\sqrt{841}}$ = 29 and
|z|2 = |5 + 2i|2 = 52 + 22 = 29
If we consider z = 0 (= 0 + 0i), then |z| = ${\sqrt{0^{2}+0^{2}}}$ = 0.
Suppose two complex numbers are: z = 2 + 3i and w = 3 + 4i.
Now, ${z\cdot w}$ = (2 + 3i)(3 + 4i) = 6 + 8i + 9i + 12i2 = -6 + 17i
${\left| z\cdot w\right|}$ = ${\sqrt{\left( -6\right) ^{2}+\left( 17\right) ^{2}}}$
= ${\sqrt{36+289}}$ = ${\sqrt{325}}$ = ${5\sqrt{13}}$
And, |z| = ${\sqrt{2^{2}+3^{2}}}$ = ${\sqrt{13}}$
|w| = ${\sqrt{3^{2}+4^{2}}}$ = ${\sqrt{25}}$ = 5
${\left| z\right| \cdot \left| w\right|}$ = ${5\sqrt{13}}$ = ${\left| z\cdot w\right|}$
For z = 2 + 3i and w = 3 + 4i,
${\left| \dfrac{z}{w}\right|}$
${\left| \dfrac{2+3i}{3+4i}\right|}$
Using The multiplicative property of modulus, we get,
${\left| \left( 2+3i\right) \dfrac{1}{\left( 3+4i\right) }\right|}$
= ${\left| 2+3i\right| \cdot \left| \dfrac{1}{3+4i}\right|}$
= ${\left| 2+3i\right| \cdot \dfrac{1}{\left| 3+4i\right| }}$
= ${\dfrac{\left| z\right| }{\left| w\right| }}$
Let us consider z = 2 + i and its conjugate ${\overline{z}}$ = 2 – i.
Now, |z| = ${\sqrt{2^{2}+1^{2}}}$ = ${\sqrt{5}}$${\left| \overline{z}\right|}$ = ${\sqrt{2^{2}+\left( -1\right) ^{2}}}$ = ${\sqrt{5}}$
Find the modulus of the complex number 8 – 6i
As we know, the absolute or modulus of z = |z| = ${\sqrt{a^{2}+b^{2}}}$
Here, |8 – 6i| = ${\sqrt{8^{2}+6^{2}}}$ = 10
Plot the complex number -4 + 5i and find its absolute value.
As we know, the absolute or modulus of z = |z| = ${\sqrt{a^{2}+b^{2}}}$
Here, |-4 + 5i| = ${\sqrt{\left( -4\right) ^{2}+5^{2}}}$ = ${\sqrt{41}}$.
Which complex number has an absolute value of 5?
a) 1 + 2i
b) 3 + 4i
c) 8 + 6i
d) 3 + 5i
As we know, the absolute value of a complex number z = a + ib is |z| = ${\sqrt{a^{2}+b^{2}}}$.
a) |1 + 2i| = ${\sqrt{1^{2}+2^{2}}}$ = ${\sqrt{1+4}}$ = ${\sqrt{5}}$
b) |3 + 4i| = ${\sqrt{3^{2}+4^{2}}}$ = ${\sqrt{9+16}}$ = ${\sqrt{25}}$ = 5
c) |8 + 6i| = ${\sqrt{8^{2}+6^{2}}}$ = ${\sqrt{64+36}}$ = ${\sqrt{100}}$ = 10
d) |3 + 5i| = ${\sqrt{3^{2}+5^{2}}}$ = ${\sqrt{9+25}}$ = ${\sqrt{34}}$
Thus, 3 + 4i has an absolute value of 5.
Last modified on February 22nd, 2024