What will be the Modulus of the complex number (5 - 5i)/(3 - 4i)?
The question asks for the modulus of the complex number given by the expression &(\frac{5 - 5i}{3 - 4i}&). We need to find the absolute value of this complex number.
For a complex number in the standard form &($z = x + iy$&), where &($x$) is the real part and &($y$) is the imaginary part, the modulus (or absolute value) is calculated using the formula:
&(|z| = \sqrt{x^2 + y^2}&)
Alternatively, for the division of two complex numbers, &($z_1 / z_2$&), the modulus can be calculated as the ratio of their moduli:
&(|z_1 / z_2| = |z_1| / |z_2|&)
Let &($z_1 = 5 - 5i$) and &($z_2 = 3 - 4i$). We can calculate the modulus of each complex number separately and then divide.
Now, divide the moduli:
&(\left|\frac{5 - 5i}{3 - 4i}\right| = \frac{|5 - 5i}}{|3 - 4i|} = \frac{\sqrt{50}}{\sqrt{25}}&)
Simplify the expression:
&(\frac{\sqrt{50}}{\sqrt{25}} = \sqrt{\frac{50}{25}} = \sqrt{2}&)
Another approach is to first simplify the complex number expression by multiplying the numerator and denominator by the conjugate of the denominator (&($3 + 4i$)):
&(\frac{5 - 5i}{3 - 4i} = \frac{(5 - 5i)(3 + 4i)}{(3 - 4i)(3 + 4i)}&)
Multiply the numerator:
&((5 - 5i)(3 + 4i) = 5(3) + 5(4i) - 5i(3) - 5i(4i)&)
&(= 15 + 20i - 15i - 20i^2&)
Since &($i^2 = -1$), we get:
&(= 15 + 5i - 20(-1) = 15 + 5i + 20 = 35 + 5i&)
Multiply the denominator (using &((a-bi)(a+bi) = a^2 + b^2)&):
&((3 - 4i)(3 + 4i) = 3^2 + (-4)^2 = 9 + 16 = 25&)
The simplified complex number is:
&(\frac{35 + 5i}{25} = \frac{35}{25} + \frac{5i}{25} = \frac{7}{5} + \frac{1}{5}i&)
Now, calculate the modulus of this simplified complex number &($z = \frac{7}{5} + \frac{1}{5}i$). Here, &($x = \frac{7}{5}$) and &($y = \frac{1}{5}$).
&(|z| = \sqrt{\left(\frac{7}{5}\right)^2 + \left(\frac{1}{5}\right)^2}&)
&(|z| = \sqrt{\frac{49}{25} + \frac{1}{25}}&)
&(|z| = \sqrt{\frac{49 + 1}{25}} = \sqrt{\frac{50}{25}}&)
&(|z| = \sqrt{2}&)
Both methods yield the same result.
The modulus of the complex number &(\frac{5 - 5i}{3 - 4i}$) is &(\sqrt{2}$).
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